Chapter 9: Ray Optics and Optical Instruments Application Questions | physics Class 12 CBSE

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Chapter 9: Ray Optics and Optical Instruments Application Questions | physics Class 12 CBSE | ByteNotes.in

Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 1

Applied Concept

A driver uses the side-view convex mirror (R = 2 m) of a parked car to watch a jogger running at 5 m/s towards the car. As the jogger approaches from 39 m to 9 m from the mirror, explain why the image appears to move faster and faster even though the jogger's speed is constant.

For a convex mirror with f = R/2 = +1 m, the mirror equation v = fu/(u - f) gives image distance v. As u changes from -39 m to -9 m, the image shifts from 39/40 m to 9/10 m — a larger fractional change for smaller u.
The speed of the image is dv/dt = [f/(u - f)]² × (du/dt). Since f/(u-f) increases as |u| decreases (jogger approaches), the image speed increases even though du/dt = 5 m/s is constant. Calculated speeds are approximately 1/280, 1/150, 1/60, and 1/10 m/s at 39, 29, 19, and 9 m respectively.
This is a practical illustration that convex mirrors compress perspective: objects near the mirror appear to move faster in the image than objects far away, which is why the mirror carries the warning 'objects in mirror are closer than they appear'.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

A tank at a swimming pool is filled with water (n = 1.33) to a depth of 2 m. A coin lies at the bottom. How deep does the coin appear to a swimmer looking straight down from above the water surface? If the tank were filled with a denser liquid of n = 1.6, by how much would the apparent depth change?

For normal (near-perpendicular) viewing, apparent depth = real depth / refractive index. For water: apparent depth = 2/1.33 ≈ 1.50 m. The coin appears to be about 1.50 m below the surface instead of the actual 2 m.
For the liquid of n = 1.6: apparent depth = 2/1.6 = 1.25 m. The coin appears even closer to the surface.
The change in apparent depth = 1.50 - 1.25 = 0.25 m, so the coin appears 25 cm closer to the surface in the denser liquid. This principle explains why swimming pools appear shallower than they are, and why a microscope focused on the bottom of a liquid-filled container must be raised when the liquid is replaced by a denser one.

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Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 3

Applied Concept

An optician prescribes spectacles of power +2.5 D for one eye and -4.0 D for the other. Identify the type of defect and the corrective lens for each eye. What are the corresponding focal lengths?

For the +2.5 D lens: P = +2.5 D, so f = 1/P = 1/2.5 = +0.4 m = +40 cm. A positive focal length means a converging (convex) lens. This eye is hypermetropic (far-sighted) and needs a convex lens to converge light onto the retina.
For the -4.0 D lens: P = -4.0 D, so f = 1/(-4.0) = -0.25 m = -25 cm. A negative focal length means a diverging (concave) lens. This eye is myopic (near-sighted) and needs a concave lens to diverge light so it focuses on the retina.
Power is the standard clinical measure because when corrective lenses are combined (e.g., for astigmatism), total power = P1 + P2 (algebraic addition), which is simpler than combining focal lengths using the reciprocal formula.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

In a medical endoscopy procedure, a bundle of optical fibres is inserted into a patient's stomach to examine it visually. Explain the physics that makes this possible, and state the requirements for the fibre material.

The optical fibre bundle works by total internal reflection. Each glass/quartz fibre has a high-refractive-index core; when light enters at a suitable angle, it strikes the core-cladding interface at an angle greater than the critical angle and is totally reflected, repeated along the full length.
Since no light is lost at each reflection (total internal reflection preserves intensity), the image signal is transmitted to the other end of the fibre with minimal loss, even through a bent fibre. A coherent bundle (fibres arranged in the same spatial order at both ends) preserves image information.
The main requirement is very low absorption of light over the fibre length. This is achieved by using highly purified materials such as quartz or silica glass; in silica fibres, over 95% of light is transmitted per kilometre, allowing clear real-time images of internal organs like the esophagus, stomach, and intestines.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

A student places a mobile phone along the principal axis of a concave mirror. Explain why the image of the phone is distorted, and whether the extent of distortion depends on where the phone is placed.

The mobile phone is an extended object along the principal axis. Different parts of the phone are at different distances from the mirror, so each part is imaged at a different image distance according to the mirror equation 1/v + 1/u = 1/f. Different parts also have different magnifications m = -v/u.
Because magnification varies continuously along the object, the image is stretched or compressed non-uniformly — the closer end to the focus has a very different magnification than the far end, resulting in a distorted image that is not geometrically similar to the object.
Yes, the extent of distortion depends strongly on the position of the phone. If the phone straddles the focus F, different parts of the phone form images on opposite sides of the mirror (some real and some virtual), greatly increasing distortion. If the phone is far from the mirror (much beyond 2f), v varies slowly with u and distortion is small.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

During a magic show, a magician submerges a glass lens (n = 1.47) in a transparent liquid and the lens becomes invisible. Identify the liquid and explain the physics. Could ordinary water serve this purpose?

For the lens to become invisible, it must not bend light at all — i.e., it must appear to have infinite focal length. From the lens maker's formula, 1/f = (n21 - 1)(1/R1 - 1/R2); for f → ∞, n21 must equal 1, meaning the refractive index of the liquid must equal that of the lens glass: nliquid = 1.47.
When n1 = n2, there is no refraction at either glass surface; the lens acts like a plane sheet and is optically indistinguishable from the surrounding liquid, making it invisible.
Water has a refractive index of approximately 1.33, which is less than 1.47, so the glass lens would still have finite focal length and would not disappear in water. The liquid could be glycerine, which has a refractive index close to 1.47, or another suitable organic liquid with the matching refractive index.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

A jeweller wants to use a simple magnifying glass to examine a gem clearly. If the jeweller's near point is at 25 cm and she needs a magnification of 5, find the required focal length. What happens to the magnification if the image is formed at infinity instead of the near point?

For image at the near point D = 25 cm: m = 1 + D/f. Given m = 5: 5 = 1 + 25/f, so 25/f = 4, giving f = 25/4 = 6.25 cm. A convex lens of focal length 6.25 cm is required.
For image at infinity: m = D/f = 25/6.25 = 4. The magnification drops by 1 (from 5 to 4) when the image is formed at infinity instead of the near point.
Despite the slight reduction in magnification, the image-at-infinity case is preferred for prolonged examination (as in gem inspection) because the eye is relaxed (focused at infinity) rather than strained (focused at the near point of 25 cm), reducing eye fatigue during extended use.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 8

Applied Concept

A beam of light converges at a point P. A concave lens of focal length 16 cm is placed 12 cm in front of P. Find the position of the final image. Is the image real or virtual?

The converging beam would meet at P in the absence of the lens. With the lens placed 12 cm before P, point P acts as a virtual object for the concave lens at u = +12 cm (positive, since it is on the same side as the transmitted light — the 'object' is behind the lens).
Using the lens formula: 1/v - 1/u = 1/f → 1/v - 1/12 = 1/(-16) → 1/v = 1/12 - 1/16 = (4 - 3)/48 = 1/48. So v = +48 cm.
The image is formed 48 cm to the right of the concave lens (on the transmitted light side), which is a real image. The concave lens has shifted the convergence point further away from the lens — a diverging lens moves the convergence point away from the lens, but since the object was virtual (converging beam), the image remains real.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 9

Applied Concept

A prism of angle 60° has a minimum deviation of 40°. Calculate its refractive index. If this prism is now placed in water (n = 1.33), would the angle of minimum deviation increase or decrease? Justify.

Using n = sin[(A + Dm)/2] / sin(A/2) = sin[(60° + 40°)/2] / sin[60°/2] = sin 50° / sin 30° = 0.766 / 0.5 = 1.532.
When the prism is placed in water, the effective refractive index of the prism with respect to water is n_eff = 1.532 / 1.33 ≈ 1.152, which is much smaller than the prism's absolute refractive index of 1.532 in air.
Since the effective refractive index decreases, the bending of light at each surface is less, and from Dm = (n_eff - 1)A (thin prism approximation), the angle of minimum deviation decreases. In water, the prism deviates light much less than in air.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 10

Applied Concept

A compound microscope has an objective of focal length 2 cm and an eyepiece of focal length 6.25 cm, with the lenses 15 cm apart. An object is placed 2.5 cm from the objective. Find the position of the final image and the total magnification.

For the objective: u = -2.5 cm, f = +2.0 cm. Using 1/v - 1/u = 1/f: 1/v = 1/2 + 1/(-2.5) = 1/2 - 2/5 = 5/10 - 4/10 = 1/10. So v = +10 cm (image is 10 cm to the right of objective).
The eyepiece is 15 cm from objective, so the intermediate image is 15 - 10 = 5 cm to the left of the eyepiece, acting as a real object at u = -5 cm. For eyepiece (f = 6.25 cm): 1/v = 1/6.25 + 1/(-5) = 0.16 - 0.20 = -0.04. So v = -25 cm — image at 25 cm to the left (virtual, at the near point).
Magnification by objective: mo = v/u = 10/(-2.5) = -4 (inverted, magnified 4×). Magnification by eyepiece: me = v/u = (-25)/(-5) = 5. Total magnification m = mo × me = (-4)(5) = -20. The negative sign confirms the final image is inverted relative to the object, with a magnitude of 20.

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Chapter 9: Ray Optics and Optical Instruments | Application Questions

A driver uses the side-view convex mirror (R = 2 m) of a parked car to watch a jogger running at 5 m/s towards the car. As the jogger approaches from 39 m to 9 m from the mirror, explain why the image appears to move faster and faster even though the jogger's speed is constant.

A tank at a swimming pool is filled with water (n = 1.33) to a depth of 2 m. A coin lies at the bottom. How deep does the coin appear to a swimmer looking straight down from above the water surface? If the tank were filled with a denser liquid of n = 1.6, by how much would the apparent depth change?

An optician prescribes spectacles of power +2.5 D for one eye and -4.0 D for the other. Identify the type of defect and the corrective lens for each eye. What are the corresponding focal lengths?

In a medical endoscopy procedure, a bundle of optical fibres is inserted into a patient's stomach to examine it visually. Explain the physics that makes this possible, and state the requirements for the fibre material.

A student places a mobile phone along the principal axis of a concave mirror. Explain why the image of the phone is distorted, and whether the extent of distortion depends on where the phone is placed.

During a magic show, a magician submerges a glass lens (n = 1.47) in a transparent liquid and the lens becomes invisible. Identify the liquid and explain the physics. Could ordinary water serve this purpose?

A jeweller wants to use a simple magnifying glass to examine a gem clearly. If the jeweller's near point is at 25 cm and she needs a magnification of 5, find the required focal length. What happens to the magnification if the image is formed at infinity instead of the near point?

A beam of light converges at a point P. A concave lens of focal length 16 cm is placed 12 cm in front of P. Find the position of the final image. Is the image real or virtual?

A prism of angle 60° has a minimum deviation of 40°. Calculate its refractive index. If this prism is now placed in water (n = 1.33), would the angle of minimum deviation increase or decrease? Justify.

A compound microscope has an objective of focal length 2 cm and an eyepiece of focal length 6.25 cm, with the lenses 15 cm apart. An object is placed 2.5 cm from the objective. Find the position of the final image and the total magnification.

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