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Ray Optics and Optical Instruments – CBSE NCERT Study Resources

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12th

12th - Physics

Ray Optics and Optical Instruments

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Overview of the Chapter

This chapter introduces the fundamental principles of Ray Optics and Optical Instruments, focusing on the behavior of light as rays and their applications in various optical devices. Topics include reflection, refraction, lenses, prisms, and optical instruments like microscopes and telescopes.

Reflection of Light

Reflection is the phenomenon where light bounces back into the same medium after striking a polished surface.

Laws of Reflection:

  • The incident ray, reflected ray, and the normal to the surface lie in the same plane.
  • The angle of incidence is equal to the angle of reflection.

Refraction of Light

Refraction is the bending of light when it passes from one medium to another due to a change in its speed.

Snell's Law: n1 sin θ1 = n2 sin θ2, where n is the refractive index.

Lenses

Lenses are transparent optical devices that refract light to form images. They can be convex (converging) or concave (diverging).

Focal length is the distance between the lens and its focus.

Lens formula: 1/f = 1/v - 1/u, where f is focal length, v is image distance, and u is object distance.

Optical Instruments

Microscope

A microscope is used to magnify small objects. It consists of two lenses: the objective lens and the eyepiece.

Telescope

A telescope is used to observe distant objects. It also uses an objective lens and an eyepiece to magnify images.

Dispersion of Light

Dispersion is the splitting of white light into its constituent colors when passing through a prism.

This occurs due to the variation in refractive index for different wavelengths of light.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
Define refractive index of a medium.
Answer:
Definition: Ratio of speed of light in vacuum to its speed in the medium.
Question 2:
What is the critical angle for total internal reflection?
Answer:
Angle of incidence in denser medium when angle of refraction becomes 90°.
Question 3:
Name the phenomenon causing twinkling of stars.
Answer:
Atmospheric refraction due to varying refractive indices of air layers.
Question 4:
State the power of a lens formula.
Answer:
Formula: P = 1/f (in meters), where f is focal length.
Question 5:
What type of lens is used to correct myopia?
Answer:
Concave lens (diverging lens).
Question 6:
Give one application of total internal reflection.
Answer:
Optical fibers in communication.
Question 7:
Define dispersion of light.
Answer:
Splitting of white light into its constituent colors due to refraction.
Question 8:
What is the magnifying power of a simple microscope?
Answer:
Ratio of angle subtended by image to angle subtended by object at eye.
Question 9:
Why does a diamond sparkle?
Answer:
Due to multiple total internal reflections and high refractive index.
Question 10:
State the lens maker's formula.
Answer:
Formula: 1/f = (μ-1)(1/R₁ - 1/R₂).
Question 11:
What is the principle behind optical fibers?
Answer:
Total internal reflection of light within the fiber core.
Question 12:
Name the defect when eye lens becomes cloudy.
Answer:
Cataract.
Question 13:
What is astigmatism?
Answer:
Eye defect due to uneven curvature of cornea causing blurred vision.
Question 14:
Give the condition for no dispersion in prism combination.
Answer:
Angular dispersion of two prisms must be equal and opposite.
Question 15:
What is the principle behind the working of a simple microscope?
Answer:

A simple microscope works on the principle of angular magnification.
It uses a single convex lens to produce a magnified virtual image of an object placed within its focal length.

Question 16:
State the condition for total internal reflection to occur.
Answer:

Total internal reflection occurs when:
1. Light travels from a denser medium to a rarer medium.
2. The angle of incidence is greater than the critical angle for the pair of media.

Question 17:
What is the function of the eyepiece in a compound microscope?
Answer:

The eyepiece in a compound microscope further magnifies the image formed by the objective lens.
It acts like a simple microscope to produce a final enlarged virtual image for the observer.

Question 18:
Why does a diamond sparkle more than a glass piece cut to the same shape?
Answer:

A diamond sparkles more due to its high refractive index (~2.42), which results in a small critical angle.
This causes most incident light to undergo total internal reflection, enhancing its brilliance.

Question 19:
Define power of a lens and state its SI unit.
Answer:

The power of a lens is the reciprocal of its focal length (in meters).
Mathematically, P = 1/f.
Its SI unit is dioptre (D).

Question 20:
What is the cause of chromatic aberration in lenses?
Answer:

Chromatic aberration occurs because a lens has different refractive indices for different wavelengths of light.
This causes dispersion, leading to colored fringes around images.

Question 21:
Why is the focal length of a convex lens less in water than in air?
Answer:

The focal length decreases in water because the relative refractive index of the lens material with respect to water is less than that with air.
This increases the lens's converging power (P = 1/f).

Question 22:
What is the significance of the aperture in optical instruments?
Answer:

The aperture determines:
1. The amount of light entering the instrument.
2. The resolving power (ability to distinguish between close objects).
Larger apertures improve brightness and resolution.

Question 23:
State the relationship between the radius of curvature (R) and focal length (f) of a spherical mirror.
Answer:

For a spherical mirror, f = R/2.
The focal length is half the radius of curvature, and both are measured from the pole of the mirror.

Question 24:
Why does the sky appear blue during the day?
Answer:

The sky appears blue due to Rayleigh scattering.
Shorter blue wavelengths of sunlight scatter more than other colors when passing through Earth's atmosphere.

Question 25:
What is the role of the objective lens in a telescope?
Answer:

The objective lens in a telescope:
1. Gathers light from distant objects.
2. Forms a real, inverted image at its focal plane.
It determines the telescope's light-gathering power and resolution.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
State the conditions for total internal reflection to occur.
Answer:

Total internal reflection occurs when:

  • Light travels from a denser medium to a rarer medium.
  • The angle of incidence is greater than the critical angle for the pair of media.

Question 2:
What is the critical angle? Give its relationship with refractive index.
Answer:

The critical angle is the minimum angle of incidence in the denser medium for which light is totally internally reflected.
It is related to the refractive index (μ) by: sin C = 1 / μ, where C is the critical angle.

Question 3:
Why does a diamond sparkle more than glass?
Answer:

A diamond sparkles more than glass due to its high refractive index (~2.42), which results in a small critical angle (~24.4°).
This causes most incident light to undergo total internal reflection, enhancing its brilliance.

Question 4:
What is the principle behind the working of an optical fiber?
Answer:

An optical fiber works on the principle of total internal reflection.
Light entering the fiber at an angle greater than the critical angle is repeatedly reflected inside the core, ensuring minimal loss of signal.

Question 5:
Differentiate between a convex lens and a concave lens in terms of their focal lengths.
Answer:

  • A convex lens has a positive focal length as it converges light rays.
  • A concave lens has a negative focal length as it diverges light rays.

Question 6:
What is dispersion of light? Name the phenomenon responsible for it.
Answer:

Dispersion of light is the splitting of white light into its constituent colors due to the variation in refractive index with wavelength.
The phenomenon responsible is refraction combined with the wavelength dependence of the refractive index.

Question 7:
Explain why the sky appears blue during the day.
Answer:

The sky appears blue due to Rayleigh scattering.
Shorter wavelengths (blue light) are scattered more by atmospheric particles than longer wavelengths, making the sky appear blue.

Question 8:
What is the function of the retina in the human eye?
Answer:

The retina acts as the light-sensitive screen in the human eye.
It contains photoreceptor cells (rods and cones) that convert light into electrical signals sent to the brain.

Question 9:
Define power of a lens. State its SI unit.
Answer:

The power of a lens is the measure of its ability to converge or diverge light.
It is the reciprocal of the focal length (in meters).
SI unit: diopter (D).

Question 10:
Why does a concave lens always form a virtual image?
Answer:

A concave lens always forms a virtual image because it diverges light rays.
The diverging rays appear to meet at a point on the same side as the object, creating a virtual, erect, and diminished image.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Explain the phenomenon of total internal reflection with a necessary diagram. State the conditions required for it to occur.
Answer:

Total internal reflection occurs when a light ray traveling from a denser medium to a rarer medium is incident at an angle greater than the critical angle, causing the ray to reflect back into the denser medium instead of refracting.

Conditions:

  • Light must travel from a denser medium to a rarer medium.
  • The angle of incidence must be greater than the critical angle for the pair of media.

Diagram: (Draw a ray incident at an angle greater than the critical angle, reflecting back into the denser medium.)

Question 2:
Derive the lens maker's formula for a thin convex lens. State the assumptions made.
Answer:

The lens maker's formula for a thin convex lens is given by:
1/f = (μ - 1)(1/R₁ - 1/R₂)
where:
f = focal length
μ = refractive index of lens material
R₁, R₂ = radii of curvature of the two surfaces.

Assumptions:

  • The lens is thin, so thickness is negligible.
  • The aperture of the lens is small.
  • The incident light is paraxial (close to the principal axis).
Question 3:
What is chromatic aberration in lenses? How can it be minimized?
Answer:

Chromatic aberration is the defect in a lens where different colors of light focus at different points due to dispersion (variation of refractive index with wavelength).

Minimization methods:

  • Use an achromatic doublet (combination of convex and concave lenses with different materials).
  • Use a narrow aperture to reduce the spread of colors.
  • Use mirrors instead of lenses (mirrors do not suffer from chromatic aberration).
Question 4:
Describe the working of a compound microscope with a ray diagram. What is its magnifying power?
Answer:

A compound microscope uses two convex lenses:

  • Objective lens (near the object) forms a real, inverted, and magnified image.
  • Eyepiece (near the eye) acts as a magnifier for the intermediate image.

Magnifying power (M) when the final image is at least distance of distinct vision (D):
M = (v₀/u₀)(1 + D/fₑ)
where v₀ = image distance for objective, u₀ = object distance, fₑ = focal length of eyepiece.

Diagram: (Draw rays from object passing through objective and eyepiece, forming a magnified virtual image.)

Question 5:
Why does the sky appear blue during the day and reddish at sunrise/sunset? Explain with Rayleigh scattering.
Answer:

The color of the sky is due to Rayleigh scattering, where shorter wavelengths (blue/violet) scatter more than longer wavelengths (red/orange).

  • Daytime: Blue light scatters dominantly in all directions, making the sky appear blue.
  • Sunrise/Sunset: Sunlight travels through a thicker layer of the atmosphere, scattering most blue light away, leaving reddish hues to reach our eyes.
Question 6:
What is a prism? Derive the expression for the angle of deviation for a prism in the minimum deviation position.
Answer:

A prism is a transparent optical element with flat, polished surfaces that refract light. The angle of deviation (δ) is the angle between the incident and emergent rays.

For minimum deviation position:
δ = (μ - 1)A
where:
μ = refractive index of prism material
A = angle of prism (apex angle).
Derivation assumes:

  • The prism is thin.
  • Light passes symmetrically through the prism.
  • Incident and emergent angles are equal.
Question 7:
Explain why the sky appears blue during the day using the concept of scattering of light.
Answer:

The sky appears blue due to the scattering of light by atmospheric molecules. Sunlight consists of seven colors (VIBGYOR), and among them, blue light has a shorter wavelength and higher frequency. According to Rayleigh's scattering law, the intensity of scattered light is inversely proportional to the fourth power of its wavelength (I ∝ 1/λ⁴). Since blue light has a shorter wavelength, it scatters more efficiently in all directions compared to other colors. This scattered blue light reaches our eyes, making the sky appear blue.

Question 8:
Define critical angle and derive its relationship with the refractive index of the medium.
Answer:

The critical angle is the angle of incidence in a denser medium for which the angle of refraction in the rarer medium becomes 90°.

Derivation:
Using Snell's law: n₁ sin i = n₂ sin r
At critical angle (i = θ_c), r = 90° and sin 90° = 1.
For denser to rarer medium: n_denser sin θ_c = n_rarer × 1
Thus, sin θ_c = n_rarer / n_denser
Since n = n_denser / n_rarer (refractive index),
sin θ_c = 1/n

Question 9:
Describe the working of a compound microscope with a neat ray diagram.
Answer:

A compound microscope uses two convex lenses: the objective lens (near the object) and the eyepiece (near the eye).

Working:
1. The object is placed just beyond the focal length of the objective lens, forming a real, inverted, and magnified image (I₁).
2. This image (I₁) acts as the object for the eyepiece, which further magnifies it to form a virtual, inverted, and highly magnified final image (I₂).

Diagram:
(Draw a labeled diagram showing object, objective lens, intermediate image, eyepiece, and final image with light rays.)

Question 10:
State the conditions for total internal reflection (TIR) to occur. Give one practical application of TIR.
Answer:

Conditions for total internal reflection:

  • Light must travel from a denser to a rarer medium.
  • The angle of incidence must be greater than the critical angle.

Application: Optical fibers use TIR to transmit light signals with minimal loss. Light entering the fiber undergoes repeated TIR at the core-cladding interface, enabling high-speed data transmission.

Question 11:
Derive the lens maker's formula for a thin convex lens.
Answer:

The lens maker's formula is derived using refraction at two spherical surfaces:

1. For the first surface (R₁): (n - 1)/R₁ = (1/v₁) - (1/u)
2. For the second surface (R₂): (1 - n)/R₂ = (1/v) - (1/v₁)
Adding both equations and assuming u ≈ ∞ (object at infinity) and v ≈ f (image at focus):
(n - 1)(1/R₁ - 1/R₂) = 1/f
Thus, 1/f = (n - 1)(1/R₁ - 1/R₂)

Question 12:
Explain how a prism produces a spectrum of white light. Why is violet light deviated more than red light?
Answer:

A prism disperses white light into its constituent colors due to refraction and wavelength-dependent refractive index.

Process:
1. White light enters the prism and refracts at the first surface.
2. Different colors (wavelengths) bend by different amounts (violet bends most, red least).
3. At the second surface, refraction further separates the colors, forming a spectrum.

Violet light deviates more because its refractive index is higher (shorter wavelength) compared to red light, as per Cauchy's relation (n = A + B/λ²).

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Explain the working of a compound microscope with a ray diagram. Derive the expression for its magnifying power when the final image is formed at the least distance of distinct vision.
Answer:
Theoretical Framework

A compound microscope uses two convex lenses: the objective (fo) and the eyepiece (fe). The object is placed just beyond fo, forming a real, inverted, and magnified image. This image acts as the object for the eyepiece, which further magnifies it.

[Diagram: Ray diagram showing light path through objective and eyepiece]
Evidence Analysis

Magnifying power (M) is given by: M = Mo × Me, where Mo is magnification by the objective and Me by the eyepiece. When the final image is at D (least distance of distinct vision), Me = (1 + D/fe). Our textbook shows Mo ≈ L/fo (L = tube length).


Critical Evaluation

This derivation assumes L >> fo and D ≈ 25 cm. Modern microscopes use infinity-corrected optics, but NCERT simplifies it for clarity.

Question 2:
Define critical angle and total internal reflection. Explain their applications in optical fibers with a diagram.
Answer:
Theoretical Framework

The critical angle is the minimum angle of incidence in a denser medium beyond which light reflects entirely (total internal reflection). It occurs when light travels from higher to lower refractive index (n1 > n2).

[Diagram: Light propagation in optical fiber]
Evidence Analysis

For optical fibers, the core (n1) is clad with lower index material (n2). Our textbook shows the condition: θi > sin⁻¹(n2/n1). This allows light to propagate through bends with minimal loss.


Future Implications

This principle enables high-speed internet (FTTH) and medical endoscopes. Current data shows fiber optics transmit >100 Tbps in lab conditions.

Question 3:
Derive the lensmaker's formula for a thin convex lens. How does it explain the chromatic aberration in lenses?
Answer:
Theoretical Framework

The lensmaker's formula relates focal length (f) to radii of curvature (R1, R2) and refractive index (μ): 1/f = (μ-1)(1/R1 - 1/R2). We studied its derivation using refraction at spherical surfaces.


Evidence Analysis

Since μ varies with wavelength (dispersion), f differs for colors, causing chromatic aberration. Our textbook shows f(red) > f(violet) due to μ(red) < μ(violet).


Critical Evaluation

This limits lens performance in telescopes. Modern solutions use achromatic doublets (e.g., crown + flint glass).

Question 4:
Explain Huygens' principle and use it to derive the laws of refraction. How does it differ from Newton's corpuscular theory?
Answer:
Theoretical Framework

Huygens' principle states that every point on a wavefront acts as a secondary source of spherical wavelets. The new wavefront is the envelope of these wavelets.

[Diagram: Wavefront propagation]
Evidence Analysis

For refraction, if v1 and v2 are speeds in media 1 and 2, our textbook shows sin i/sin r = v1/v2 = μ (Snell's law). Newton's theory incorrectly predicted v2 > v1 in denser media.


Critical Evaluation

Huygens' principle was later validated by Foucault's speed measurements (1850), proving light slows in denser media.

Question 5:
Describe the construction and working of a astronomical telescope with a ray diagram. Why can't we use it for terrestrial viewing?
Answer:
Theoretical Framework

An astronomical telescope uses two convex lenses: objective (long fo) and eyepiece (short fe). Parallel rays from distant objects converge at the focal plane, forming an inverted image.

[Diagram: Ray paths in telescope]
Evidence Analysis

Magnifying power M = fo/fe. Our textbook shows terrestrial telescopes add an erecting lens, but this reduces brightness and increases length.


Future Implications

Modern telescopes like Hubble use mirrors to avoid chromatic aberration. Current data shows JWST has 6.5m primary mirror.

Question 6:
What is optical path length? How does it explain the formation of interference fringes in Young's double-slit experiment?
Answer:
Theoretical Framework

Optical path length (OPL) is the product of geometric path length and refractive index (μ×d). It accounts for phase changes in different media.


Evidence Analysis

In Young's experiment, OPL difference Δ = S2P - S1P ≈ yd/D (y = fringe position, D = screen distance). Our textbook shows bright fringes occur when Δ = nλ.

[Diagram: Double-slit setup]
Critical Evaluation

This validates light's wave nature. Current quantum optics experiments show single photons also form interference patterns.

Question 7:
Explain spherical aberration and coma in lenses. How are they minimized in modern optical instruments?
Answer:
Theoretical Framework

Spherical aberration occurs when marginal and paraxial rays focus at different points. Coma causes off-axis point sources to appear comet-shaped.


Evidence Analysis

Our textbook shows these arise from the spherical shape of lens surfaces. Modern solutions include:

  • Aspheric lenses
  • Multi-element designs


Future Implications

Space telescopes like Euclid (2023) use freeform optics to achieve <0.1 arcsecond resolution.

Question 8:
Derive the relation between focal length (f) and radius of curvature (R) for a spherical mirror. Why can't this be directly applied to lenses?
Answer:
Theoretical Framework

For mirrors, f = R/2 is derived using the paraxial approximation (small angles). This comes from the mirror equation 1/v + 1/u = 1/f.


Evidence Analysis

Lenses have two refracting surfaces and depend on μ. Our textbook shows lensmaker's formula requires (μ-1) term and accounts for both curvatures.


Critical Evaluation

Mirror formula is simpler as it involves reflection only. This makes mirrors preferable in large telescopes (e.g., VLT's 8.2m mirrors).

Question 9:
Describe the phenomenon of polarization by scattering with an example. How is this used in polarized sunglasses?
Answer:
Theoretical Framework

When unpolarized light scatters off molecules (Rayleigh scattering), the perpendicular component dominates, causing polarization.


Evidence Analysis

Our textbook shows this explains the blue sky and polarized light at 90° to the sun. Sunglasses use Polaroid filters to block horizontally polarized glare from roads/water.

[Diagram: Scattering polarization]
Future Implications

Liquid crystal displays (LCDs) also use polarization, with current smartphones achieving >1000:1 contrast ratio.

Question 10:
Explain the formation of rainbow using concepts of refraction, dispersion, and total internal reflection. Why is the secondary rainbow fainter?
Answer:
Theoretical Framework

A rainbow forms when sunlight undergoes:

  • Refraction + dispersion in raindrops
  • Total internal reflection
  • Final refraction

[Diagram: Light path in raindrop]
Evidence Analysis

Primary rainbow (42° radius) has one internal reflection. Secondary rainbow (51°) has two reflections, causing:

  • Color reversal
  • Higher light loss (50% more)


Critical Evaluation

Our textbook shows this explains why secondary rainbows are 1/10th as bright as primary.

Question 11:
Explain the working of a compound microscope with a neat ray diagram. Derive the expression for its magnifying power when the final image is formed at the least distance of distinct vision.
Answer:

A compound microscope is an optical instrument used to magnify tiny objects. It consists of two convex lenses: the objective lens (near the object) and the eyepiece (near the eye).

Working Principle:
1. The object is placed just beyond the focal length of the objective lens, forming a real, inverted, and magnified image (I₁).
2. This image acts as the object for the eyepiece, which further magnifies it to form a virtual, inverted, and highly magnified final image (I₂) at the least distance of distinct vision (D = 25 cm).

Ray Diagram: (Draw a labeled diagram showing the path of rays through both lenses, object position, intermediate image, and final image formation at D.)

Derivation of Magnifying Power (m):
Magnifying power is given by: m = m₀ × mₑ
where:
m₀ = magnification by objective = v₀/u₀ ≈ L/f₀ (since u₀ ≈ f₀ and v₀ ≈ L, the tube length)
mₑ = magnification by eyepiece = (1 + D/fₑ) (for image at D)
Thus, m = (L/f₀) × (1 + D/fₑ)

Note: The negative sign indicates an inverted image. For maximum magnification, both lenses should have small focal lengths.

Question 12:
Explain the working of a compound microscope with a neat ray diagram. Derive the expression for its magnifying power when the final image is formed at the least distance of distinct vision.
Answer:

A compound microscope is an optical instrument used to magnify tiny objects. It consists of two convex lenses: the objective lens (near the object) and the eyepiece (near the eye).

Working:
1. The object is placed just beyond the focal length of the objective lens.
2. The objective lens forms a real, inverted, and magnified image (called the intermediate image).
3. This intermediate image acts as the object for the eyepiece, which further magnifies it to form a virtual, inverted, and highly magnified final image.

Ray Diagram: (Diagram should show object, objective lens, intermediate image, eyepiece, and final image at D.)

Magnifying Power (m) when image is at D:
Magnifying power is the ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the eye when placed at D.
m = mo × me
where mo = (vo/uo) ≈ (L/fo) (since uo ≈ fo and vo ≈ L)
and me = (1 + D/fe)
Thus, m = (L/fo)(1 + D/fe)

Note: L is the tube length (distance between the second focal point of the objective and the first focal point of the eyepiece).

Question 13:
What is optical aberration? Explain chromatic aberration in lenses and describe how it can be minimized using an achromatic doublet.
Answer:

Optical aberration refers to defects in image formation caused by limitations in lens design or material properties, leading to blurred or distorted images.

Chromatic Aberration:
It occurs because a lens refracts different colors (wavelengths) of light by different amounts (dispersion). Shorter wavelengths (violet/blue) bend more than longer wavelengths (red), causing:
1. Longitudinal Chromatic Aberration: Different colors focus at different points along the axis.
2. Lateral Chromatic Aberration: Variation in image size for different colors.

Achromatic Doublet Solution:
An achromatic doublet combines a convex lens (crown glass) and a concave lens (flint glass) to cancel dispersion:
1. The convex lens converges light but disperses colors.
2. The concave lens diverges light but compensates for dispersion.
3. Together, they bring two primary colors (e.g., red and blue) to the same focus, minimizing chromatic aberration.

Key Point: The doublet ensures the focal length remains nearly constant across wavelengths, improving image clarity. This principle is used in high-quality cameras and telescopes.

Question 14:
What is optical aberration? Discuss chromatic aberration in detail, its causes, and methods to minimize it in lenses.
Answer:

Optical aberration refers to defects in image formation by lenses/mirrors due to limitations in their design or material properties.

Chromatic Aberration:
It is the failure of a lens to focus all colors (wavelengths) of light at the same point, causing colored fringes around images.

Causes:

  • Dispersion: Different wavelengths (colors) travel at different speeds in the lens material, causing varying refractive indices.
  • Result: Violet light bends more than red light, leading to separate focal points for different colors.

Minimization Methods:

  • Achromatic Doublet: Combines convex (crown glass) and concave (flint glass) lenses with different dispersive powers to cancel chromatic effects.
  • Use of Narrow Aperture: Reduces the amount of dispersed light entering the system.
  • Mirrors: Since reflection doesn't depend on wavelength, mirrors don't suffer chromatic aberration.

Application: High-quality cameras and telescopes use achromatic lenses to produce clear, color-accurate images.

Question 15:
Explain the working principle of a compound microscope with a neat ray diagram. Derive an expression for its magnifying power when the final image is formed at the least distance of distinct vision.
Answer:

A compound microscope is an optical instrument used to magnify tiny objects that are not visible to the naked eye. It consists of two convex lenses: the objective lens (near the object) and the eyepiece (near the eye).

Working Principle:
The objective lens forms a real, inverted, and magnified image of the object within its focal length. This image acts as an object for the eyepiece, which further magnifies it to form a virtual, inverted, and highly enlarged final image at the least distance of distinct vision (D = 25 cm).

Ray Diagram:
1. Draw the object AB placed just beyond the focus of the objective lens (Fo).
2. The objective lens forms an intermediate image A'B' (real, inverted, and magnified).
3. The eyepiece is adjusted so that A'B' lies within its focal length (Fe).
4. The eyepiece acts as a magnifying glass, producing the final virtual image A''B'' at D.

Magnifying Power (M) Derivation:
M = Angular size of final image / Angular size of object when placed at D
For the objective lens: Lateral magnification (mo) = A'B'/AB ≈ L/fo (where L is the tube length, fo is the focal length of the objective).
For the eyepiece: Angular magnification (me) = (1 + D/fe).
Thus, total magnifying power (M) = mo × me = (L/fo) × (1 + D/fe).

Note: The negative sign indicates an inverted image. Higher magnification is achieved by using lenses with shorter focal lengths.

Question 16:
Describe the phenomenon of total internal reflection (TIR) with necessary conditions. Explain its application in optical fibers with a diagram.
Answer:

Total internal reflection (TIR) is the phenomenon where a light ray traveling from a denser medium to a rarer medium is completely reflected back into the denser medium at the interface, provided:

  • The angle of incidence (i) exceeds the critical angle (c).
  • Light travels from a denser medium (e.g., glass) to a rarer medium (e.g., air).

Critical Angle (c): It is the angle of incidence for which the angle of refraction becomes 90°. It is given by: sin c = 1/μ, where μ is the refractive index of the denser medium.

Application in Optical Fibers:
Optical fibers are thin, flexible strands of glass/plastic that transmit light via TIR. The fiber consists of:

  • Core: Inner denser medium (high μ).
  • Cladding: Outer rarer medium (low μ).

Diagram:
1. Draw a cylindrical optical fiber with core and cladding layers.
2. Show a light ray entering the core at an angle > critical angle.
3. The ray undergoes multiple TIRs at the core-cladding interface, propagating through the fiber without escaping.

Advantages:
Optical fibers are used in telecommunications due to:
- High bandwidth.
- Low signal loss.
- Immunity to electromagnetic interference.

Question 17:
What is total internal reflection? State the conditions for it. Explain how this phenomenon is used in optical fibers with a diagram.
Answer:

Total internal reflection (TIR) is the phenomenon where a light ray traveling from a denser medium to a rarer medium is completely reflected back into the denser medium when the angle of incidence exceeds the critical angle.

Conditions for TIR:
1. Light must travel from a denser to a rarer medium.
2. The angle of incidence must be greater than the critical angle for the pair of media.

Application in Optical Fibers:
Optical fibers are thin, flexible strands of glass or plastic that use TIR to transmit light signals over long distances with minimal loss.
1. The fiber consists of a core (denser medium) surrounded by a cladding (rarer medium).
2. Light entering the core at an angle greater than the critical angle undergoes TIR at the core-cladding interface.
3. This repeated TIR allows the light to propagate through the fiber with little energy loss.

Diagram: (Diagram should show light rays entering the fiber core, undergoing multiple TIRs, and exiting the other end.)

Advantages: Optical fibers are used in telecommunications due to their high bandwidth, immunity to electromagnetic interference, and low signal attenuation.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
A student observes that a convex lens forms a virtual, erect, and magnified image when the object is placed within its focal length. However, when the same lens is immersed in water, the image characteristics change. Analyze this phenomenon using the lens maker's formula and explain why the behavior differs in water.
Answer:
Case Deconstruction

We studied that a convex lens in air converges light due to its refractive index being higher than air. When the object is within the focal length, a virtual image forms.


Theoretical Application

The lens maker's formula shows that focal length depends on the refractive index of the surrounding medium. In water, the relative refractive index decreases, reducing the lens's converging power, which may alter image characteristics.


Critical Evaluation
  • Example 1: A lens with μ=1.5 in air (μ=1) has higher power than in water (μ=1.33).
  • Example 2: Underwater cameras use specialized lenses to compensate for this effect.
Question 2:
In an experiment, a prism disperses white light into its constituent colors, but a glass slab does not. Using the concepts of refraction and angular dispersion, justify why the prism causes dispersion while the slab does not.
Answer:
Case Deconstruction

Our textbook shows that dispersion occurs due to wavelength-dependent refraction. A prism's angled surfaces cause light rays to bend at different angles for different colors.


Theoretical Application

The angular dispersion arises because the prism's non-parallel surfaces allow colors to separate. In a glass slab, the entry and exit refractions cancel out, resulting in no net dispersion.


Critical Evaluation
  • Example 1: A 60° prism creates a visible spectrum, while a slab only shifts the light laterally.
  • Example 2: Rainbow formation relies on prism-like water droplets, not flat surfaces.
Question 3:
A telescope uses an objective lens of focal length 100 cm and an eyepiece of 5 cm. Calculate its magnifying power. Critically analyze how this magnification would change if the eyepiece is replaced with one of 2.5 cm focal length, keeping the objective unchanged.
Answer:
Case Deconstruction

We know that magnifying power (M) of a telescope is given by M = fo/fe. Here, fo = 100 cm, fe = 5 cm, so M = 20.


Theoretical Application

If fe reduces to 2.5 cm, M doubles to 40. However, excessive magnification may reduce brightness and increase aberrations.


Critical Evaluation
  • Example 1: Astronomical telescopes prioritize light gathering over extreme magnification.
  • Example 2: High-power microscopes face similar trade-offs between M and image quality.
Question 4:
When a thin film of oil on water shows colorful patterns, it demonstrates interference. Explain how varying film thickness affects the observed colors, linking this to the path difference and constructive interference conditions.
Answer:
Case Deconstruction

We studied that thin-film interference occurs due to light waves reflecting off the top and bottom surfaces, creating path differences.


Theoretical Application

For constructive interference, the path difference must equal integer wavelengths. Thicker films increase the path difference, shifting the observed color (e.g., red → blue as thickness decreases).


Critical Evaluation
  • Example 1: Soap bubbles show color bands corresponding to thickness variations.
  • Example 2: Anti-reflective coatings use precisely controlled thickness to cancel specific wavelengths.
Question 5:
A student observes that a convex lens forms a virtual, erect, and magnified image when an object is placed between the lens and its focal point. Using this observation, explain how a simple microscope works. Also, compare its magnifying power with a compound microscope.
Answer:
Case Deconstruction

We studied that a convex lens forms a virtual, erect, and magnified image when the object is placed within its focal length. This principle is used in a simple microscope.

Theoretical Application
  • A simple microscope uses a single convex lens to magnify small objects by placing them within the focal length.
  • The magnifying power (M) is given by M = 1 + (D/f), where D is the least distance of distinct vision (25 cm) and f is the focal length.
Critical Evaluation

Compared to a compound microscope, which uses two lenses (objective and eyepiece), a simple microscope has lower magnification but is more portable and easier to use.

Question 6:
In an experiment, a student uses a prism to observe the dispersion of light. Explain why violet light deviates more than red light. Also, discuss how this phenomenon is utilized in optical instruments like spectrometers.
Answer:
Case Deconstruction

We learned that dispersion occurs because different wavelengths of light travel at different speeds in a medium, causing varying deviations.

Theoretical Application
  • Violet light has a shorter wavelength and higher refractive index, so it deviates more than red light.
  • In spectrometers, prisms separate light into its constituent colors, allowing analysis of spectral lines.
Critical Evaluation

This principle is crucial in identifying elements in stars or chemical samples. However, chromatic aberration in lenses is an unwanted effect of dispersion.

Question 7:
A telescope has an objective lens of focal length 100 cm and an eyepiece of 5 cm. Calculate its magnifying power in normal adjustment. Also, justify why astronomical telescopes produce inverted images.
Answer:
Case Deconstruction

Our textbook shows that the magnifying power (M) of a telescope is given by M = fo/fe, where fo and fe are focal lengths of the objective and eyepiece.

Theoretical Application
  • Here, M = 100/5 = 20.
  • Astronomical telescopes use two convex lenses, causing the image to invert. This doesn’t affect celestial observations.
Critical Evaluation

While inversion is irrelevant for stars, terrestrial telescopes use additional lenses to erect the image, adding complexity.

Question 8:
A student notices that a mirage appears on a hot road. Explain this phenomenon using total internal reflection and refractive index gradients. How does this differ from the twinkling of stars?
Answer:
Case Deconstruction

Mirages occur due to total internal reflection caused by varying refractive indices in air layers near hot surfaces.

Theoretical Application
  • Hot air near the road has a lower refractive index, bending light upward, creating a virtual image of the sky.
  • Star twinkling results from atmospheric refraction due to turbulence, causing rapid changes in apparent position.
Critical Evaluation

While both involve refraction, mirages are localized and static, whereas twinkling is dynamic and affects distant point sources.

Question 9:
A student observes that a convex lens forms a virtual, erect, and magnified image when the object is placed between the focus and the lens. Using the lens formula, explain why this occurs and compare it with the image formed by a concave lens.
Answer:
Case Deconstruction

We studied that for a convex lens, when the object is placed between the focus (F) and the lens, the image is virtual, erect, and magnified. The lens formula (1/f = 1/v - 1/u) confirms this as 'u' is negative and 'v' becomes positive, indicating a virtual image.

Theoretical Application
  • Convex lens: Virtual image forms due to diverging rays appearing to meet.
  • Concave lens: Always forms a virtual, erect, and diminished image regardless of object position.
Critical Evaluation

Our textbook shows that convex lenses are used in magnifying glasses, while concave lenses correct myopia. Both follow the lens formula but produce different image types.

Question 10:
In an experiment, a prism disperses white light into its constituent colors. Analyze the role of refractive index and angle of deviation in this phenomenon. How does this differ from dispersion in a rainbow?
Answer:
Case Deconstruction

We learned that a prism disperses light due to the variation in refractive index with wavelength. Shorter wavelengths (violet) deviate more than longer ones (red), causing dispersion.

Theoretical Application
  • Refractive index: Higher for violet light, causing greater bending.
  • Angle of deviation: Minimum for a specific wavelength, leading to a spectrum.
Critical Evaluation

In a rainbow, dispersion occurs due to refraction and reflection in water droplets, not a prism. Our textbook shows both phenomena rely on wavelength-dependent refraction but differ in their mechanisms.

Question 11:
A telescope uses an objective lens and an eyepiece to magnify distant objects. Derive the expression for its angular magnification and explain why the objective lens has a larger aperture than the eyepiece.
Answer:
Case Deconstruction

We studied that the angular magnification (M) of a telescope is given by M = fo/fe, where fo and fe are focal lengths of the objective and eyepiece.

Theoretical Application
  • Objective lens: Larger aperture collects more light, improving resolution.
  • Eyepiece: Smaller aperture focuses the image for the eye.
Critical Evaluation

Our textbook shows that a larger objective aperture reduces diffraction effects, enhancing clarity. Examples include astronomical telescopes and terrestrial binoculars.

Question 12:
A compound microscope uses two convex lenses to magnify tiny objects. Compare its working principle with a simple microscope and justify why the former provides higher magnification.
Answer:
Case Deconstruction

We learned that a compound microscope uses an objective and eyepiece lens, while a simple microscope uses a single lens. The compound microscope's magnification is the product of both lenses' magnifications.

Theoretical Application
  • Compound microscope: M = mo × me, where mo and me are magnifications of the objective and eyepiece.
  • Simple microscope: Limited by the focal length of a single lens.
Critical Evaluation

Our textbook shows that compound microscopes are used in labs for high-resolution imaging, while simple microscopes are portable but less powerful.

Question 13:
A student observes that a convex lens forms a virtual, erect, and magnified image when the object is placed between the lens and its focal point. However, when the object is moved beyond the focal point, the image becomes real and inverted. Explain this phenomenon using the lens formula and ray diagrams.
Answer:
Case Deconstruction

We studied that a convex lens converges light rays. When the object is within the focal length, rays diverge, forming a virtual image. Beyond the focal point, rays converge to a real image.


Theoretical Application
  • Using the lens formula: 1/f = 1/v - 1/u, for u < f, v is negative (virtual image).
  • For u > f, v is positive (real image).

Critical Evaluation

Ray diagrams confirm this: [Diagram: Two cases showing virtual and real image formation]. Our textbook shows similar examples for concave mirrors.

Question 14:
In an experiment, a prism disperses white light into its constituent colors, with violet light deviating the most and red the least. Analyze why this happens and how the refractive index varies with wavelength.
Answer:
Case Deconstruction

We learned that dispersion occurs due to wavelength-dependent refractive index. Shorter wavelengths (violet) bend more than longer ones (red).


Theoretical Application
  • Refractive index (n) is higher for violet light (λ ≈ 400 nm) than red (λ ≈ 700 nm).
  • Cauchy’s formula: n = A + B/λ², explains this variation.

Critical Evaluation

Our textbook shows how this principle is used in spectrometers. [Diagram: Prism dispersing light into a spectrum].

Question 15:
A telescope has an objective lens of focal length 100 cm and an eyepiece of 5 cm. Calculate its magnifying power and explain how adjusting the eyepiece affects the image.
Answer:
Case Deconstruction

We studied that telescope magnification (M) = fo/fe. Here, M = 100/5 = 20.


Theoretical Application
  • Increasing fe reduces M but widens the field of view.
  • Decreasing fe increases M but may reduce brightness.

Critical Evaluation

Our textbook shows examples like astronomical vs. terrestrial telescopes. [Diagram: Ray paths in a refracting telescope].

Question 16:
A compound microscope uses an objective lens (fo = 1 cm) and eyepiece (fe = 5 cm) separated by 20 cm. Derive its magnifying power and discuss why the object must be placed close to the focal point of the objective.
Answer:
Case Deconstruction

We know M = (vo/uo) × (D/fe). For maximum magnification, uo ≈ fo.


Theoretical Application
  • Using lens formula for objective: 1/fo = 1/vo - 1/uo.
  • For uo ≈ fo, vo ≈ L (tube length).

Critical Evaluation

Our textbook shows how this setup maximizes angular magnification. [Diagram: Microscope ray paths].

Question 17:
A student is observing the formation of images using a convex lens of focal length 20 cm. The object is placed at different distances from the lens. Based on the given scenario, answer the following:

(a) What type of image is formed when the object is placed at 30 cm from the lens?
(b) Calculate the position of the image if the object is placed at 10 cm from the lens.
Answer:

(a) When the object is placed at 30 cm from the convex lens (which is between 2F and F), the image formed is real, inverted, and diminished. This is because the object distance is greater than the focal length but less than twice the focal length.


(b) To calculate the image position when the object is placed at 10 cm (which is less than the focal length), we use the lens formula:

1/f = 1/v - 1/u

Given: f = 20 cm, u = -10 cm (object distance is negative as per sign convention).

Substituting the values:

1/20 = 1/v - 1/(-10)

1/20 = 1/v + 1/10

1/v = 1/20 - 1/10 = (1 - 2)/20 = -1/20

v = -20 cm

The negative sign indicates the image is formed on the same side as the object, which means it is a virtual image. The image is also erect and magnified in this case.

Question 18:
A ray of light passes through a glass prism of refractive index 1.5 and angle of prism 60°. The angle of incidence is 45°. Based on this information, answer the following:

(a) Calculate the angle of deviation for the given setup.
(b) Explain how the angle of deviation changes if the angle of incidence is increased gradually.
Answer:

(a) To calculate the angle of deviation, we use the formula for deviation in a prism:

δ = i + e - A

Where:

δ = angle of deviation
i = angle of incidence (45°)
e = angle of emergence (to be calculated)
A = angle of prism (60°)

First, we find the angle of refraction (r) at the first surface using Snell's law:

sin i / sin r = μ (refractive index)

sin 45° / sin r = 1.5

sin r = sin 45° / 1.5 ≈ 0.471

r ≈ sin⁻¹(0.471) ≈ 28°

Now, the angle of incidence at the second surface (r') is:

r' = A - r = 60° - 28° = 32°

Using Snell's law again for the second surface:

sin e / sin r' = 1/μ (since light is going from glass to air)

sin e = 1.5 * sin 32° ≈ 0.794

e ≈ sin⁻¹(0.794) ≈ 52.6°

Now, the angle of deviation is:

δ = 45° + 52.6° - 60° ≈ 37.6°


(b) As the angle of incidence increases gradually, the angle of deviation first decreases to a minimum value (called minimum deviation) and then increases again. This happens because the path of light becomes symmetric at the angle of minimum deviation, where the angle of incidence equals the angle of emergence. Beyond this point, further increase in the angle of incidence causes the deviation to increase.

Question 19:
A student is observing the formation of images using a convex lens of focal length 15 cm. The object is placed at a distance of 30 cm from the lens.

Based on this scenario, answer the following:

  • Calculate the position of the image formed.
  • State the nature (real/virtual) and size of the image relative to the object.
  • Draw a ray diagram to illustrate the image formation.
Answer:

To find the position of the image, we use the lens formula:
1/f = 1/v - 1/u
Given: f = 15 cm (convex lens), u = -30 cm (object distance, negative as per sign convention).
Substituting the values:
1/15 = 1/v - 1/(-30)
1/v = 1/15 - 1/30 = (2 - 1)/30 = 1/30
v = 30 cm.
The image is formed at 30 cm on the other side of the lens.

The image is real and inverted because the object is placed beyond the focal length (u > f). The magnification (m) is given by:
m = v/u = 30/(-30) = -1
The negative sign indicates inversion, and the magnitude (1) shows the image is of the same size as the object.

Ray Diagram:
1. Draw the convex lens with its principal axis.
2. Place the object (an arrow) at 30 cm on the left side of the lens.
3. Draw a ray parallel to the principal axis, which refracts through the focal point on the right.
4. Draw a ray passing through the optical center, which goes straight.
5. The intersection of these rays on the right side at 30 cm forms the real, inverted image of the same size.

Question 20:
A prism with an angle of 60° produces a minimum deviation of 30° for a monochromatic light.

Based on this information, answer the following:

  • Calculate the refractive index of the prism material.
  • Explain why the deviation is minimum in this case.
  • What happens to the deviation angle if the wavelength of light is increased?
Answer:

The refractive index (μ) of the prism material is calculated using the formula for minimum deviation (Dm):
μ = sin[(A + Dm)/2] / sin(A/2)
Given: A = 60°, Dm = 30°.
Substituting the values:
μ = sin[(60° + 30°)/2] / sin(60°/2) = sin(45°)/sin(30°) = (1/√2)/(1/2) = √2 ≈ 1.414.

The deviation is minimum when the light passes symmetrically through the prism, i.e., the angle of incidence equals the angle of emergence. This results in the least bending of light, and the path of light inside the prism becomes parallel to the base.

If the wavelength of light is increased, the refractive index decreases (due to dispersion). Since deviation depends on μ, the angle of deviation will decrease for longer wavelengths (e.g., red light) compared to shorter wavelengths (e.g., violet light).

Question 21:
A student observes that a convex lens forms a real, inverted, and diminished image of an object placed beyond 2F. The student then moves the object between F and 2F.

Based on this scenario, answer the following:

  • What change occurs in the nature and size of the image?
  • Justify your answer using the lens formula and ray diagram.
Answer:

When the object is moved from beyond 2F to between F and 2F:

  • The image changes from diminished to magnified while remaining real and inverted.

Justification using lens formula:


For a convex lens, the lens formula is:
1/f = 1/v - 1/u
When the object is beyond 2F, u > 2f, so v lies between f and 2f, giving a diminished image.
When the object is between F and 2F, f < u < 2f, so v > 2f, resulting in a magnified image.

Ray diagram explanation:


Draw two rays from the object:
1. A ray parallel to the principal axis refracts through the focus.
2. A ray passing through the optical center goes undeviated.
For u > 2f, the image is smaller and between f and 2f.
For f < u < 2f, the image is larger and beyond 2f.
Question 22:
A compound microscope consists of an objective lens of focal length 2 cm and an eyepiece of focal length 5 cm. The lenses are placed 20 cm apart. An object is placed 2.5 cm from the objective lens.

Based on this setup, answer:

  • Calculate the position of the final image formed by the microscope.
  • Explain why the eyepiece is also called the 'magnifying lens'.
Answer:

Position of the final image:


For the objective lens:
Given: fo = 2 cm, uo = -2.5 cm
Using lens formula: 1/fo = 1/vo - 1/uo
Substitute values:
1/2 = 1/vo - 1/(-2.5)
1/vo = 1/2 - 1/2.5 = 0.1
Thus, vo = 10 cm (real image by objective).

For the eyepiece:
Image by objective acts as object for eyepiece.
Distance between lenses = 20 cm, so ue = 20 - 10 = 10 cm
Given: fe = 5 cm
Using lens formula: 1/fe = 1/ve - 1/ue
1/5 = 1/ve - 1/10
1/ve = 1/5 + 1/10 = 0.3
Thus, ve ≈ 3.33 cm (virtual final image).

Eyepiece as magnifying lens:
The eyepiece further magnifies the intermediate image formed by the objective. Since it operates with the image near its focus, it produces a highly magnified virtual image, enhancing angular magnification for the observer.

Question 23:
A student observes that a convex lens forms a real, inverted, and diminished image of an object placed beyond 2F. However, when the object is placed between F and 2F, the image becomes real, inverted, and magnified. Explain the phenomenon using the lens formula and ray diagrams. Also, state one practical application of this behavior in optical instruments.
Answer:

The behavior of the convex lens can be explained using the lens formula:
1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance.

When the object is placed beyond 2F:
- The image is formed between F and 2F on the other side of the lens.
- Since u > 2f, the image is real, inverted, and diminished.

When the object is placed between F and 2F:
- The image is formed beyond 2F on the other side.
- Since f < u < 2f, the image is real, inverted, and magnified.

Ray diagrams for both cases should show:
- A ray parallel to the principal axis refracting through the focal point.
- A ray passing through the optical center continuing undeviated.
- The intersection point of these rays determines the image position.

One practical application is in projectors, where a convex lens is used to form a magnified real image of a slide or film on a screen.

Question 24:
A student uses a convex lens to focus sunlight onto a piece of paper, causing it to burn. The lens has a focal length of 20 cm. Calculate the distance at which the paper should be placed from the lens to achieve maximum burning effect. Explain why this happens and discuss the role of the lens in this phenomenon.
Answer:

To achieve maximum burning effect, the paper should be placed at the focal point of the convex lens, which is 20 cm from the lens.

Explanation:
- Sunlight consists of parallel rays of light coming from a distant object (the Sun).
- A convex lens converges these parallel rays to a point called the focus.
- At the focal point, the light energy is concentrated, producing intense heat that can burn the paper.

Role of the lens:
- The convex lens refracts the incoming parallel rays, causing them to converge.
- The converging action increases the energy density at the focal point, leading to the burning effect.

This phenomenon demonstrates the principle of image formation by a convex lens for distant objects, where the image is formed at the focal plane.

Question 25:
A student is observing the formation of images using a convex lens of focal length 15 cm. The object is placed at a distance of 30 cm from the lens.

(i) Calculate the position of the image formed.
(ii) Determine the nature and size of the image if the object height is 4 cm.
(iii) Draw a labeled ray diagram to illustrate the image formation.

Answer:

(i) Using the lens formula:
1/f = 1/v - 1/u
Given: f = +15 cm (convex lens), u = -30 cm (object distance)
1/15 = 1/v - 1/(-30)
1/v = 1/15 - 1/30 = (2-1)/30 = 1/30
v = +30 cm
The image is formed 30 cm from the lens on the opposite side of the object.

(ii) Magnification (m) = v/u = 30/(-30) = -1
Since m is negative, the image is real and inverted.
Size of image = |m| × object height = 1 × 4 cm = 4 cm. The image is of the same size as the object.

(iii) Ray diagram:
- Draw the principal axis and mark the lens center (O).
- Place the object (AB) perpendicular to the axis at 30 cm from O.
- Draw a ray parallel to the axis from A, refracting through the focal point (F) on the other side.
- Draw a ray passing through the optical center (O) without deviation.
- The intersection point gives the real, inverted image (A'B') at 30 cm.

Question 26:
A prism with an angle of 60° produces a minimum deviation of 30° for a monochromatic light.

(i) Calculate the refractive index of the prism material.
(ii) If the angle of incidence is increased beyond the angle of minimum deviation, how will the angle of deviation change? Justify.
(iii) State one application of such a prism in optical instruments.

Answer:

(i) Using the prism formula for minimum deviation:
μ = sin[(A + δm)/2] / sin(A/2)
Given: A = 60°, δm = 30°
μ = sin[(60° + 30°)/2] / sin(60°/2) = sin(45°)/sin(30°) = (1/√2)/(1/2) = √2 ≈ 1.414.

(ii) The angle of deviation increases if the angle of incidence is increased beyond the angle of minimum deviation.
Justification: The graph of deviation vs. incidence angle is a U-shaped curve. At δm, the curve is at its minimum. Any increase in incidence angle on either side leads to higher deviation.

(iii) Application: Such prisms are used in spectrometers to study the dispersion of light into its constituent wavelengths, helping analyze light sources or material properties.

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