NCERT Exemplar Ray Optics & Optical Instruments - Physics Class 12 - NEET

Multiple-Choice Questions

Q.1. A ray of light incident at an angle θ on a refracting face of a prism emerges from the other face normally. If the angle of the prism is 5° and the prism is made of a material of refractive index 1.5, the angle of incidence is 
(a) 7.5°. 
(b) 5°. 
(c) 15°. 
(d) 2.5°.

Correct Answer is Option (a)

Given:

A = 5°; μ = 1.5.

Reasoning and steps:

If the emergent ray is normal to the second face then the angle of emergence in air i₂ = 0° and the corresponding refracted angle inside the prism at the second face r₂ = 0°.

For a prism, r₁ + r₂ = A. Therefore r₁ = A - r₂ = 5° - 0° = 5°.

Apply Snell's law at the first face (air → prism): sin i₁ = μ sin r₁.

Compute sin i₁ = 1.5 × sin 5° ≈ 1.5 × 0.08716 ≈ 0.13074.

Thus i₁ = arcsin(0.13074) ≈ 7.5°.

Hence option (a) is correct.

Q.2. A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is
(a) Blue.
(b) Green.
(c) Violet.
(d) Red.

Correct Answer is Option (d)

Explanation:

The frequency f of each colour remains unchanged on refraction; the speed v in the medium is related to wavelength λ by v = fλ. In a given medium the refractive index μ = c / v, so larger wavelength λ corresponds to larger speed v in the medium (for the same f). Among visible colours, red has the largest wavelength and therefore the largest speed in glass. A short pulse will spread because different colours travel at different speeds; hence red emerges first from the slab.

Q.3. An object approaches a convergent lens from the left of the lens with a uniform speed 5 m/s and stops at the focus. The image
(a) Moves away from the lens with a uniform speed 5 m/s.
(b) Moves away from the lens with uniform acceleration.
(c) Moves away from the lens with a non-uniform acceleration.
(d) Moves towards the lens with a non-uniform acceleration.

Correct Answer is Option (c)

Explanation:

Let the object approach the lens with constant speed. The lens formula 1/v + 1/u = 1/f relates object distance u and image distance v. As u changes uniformly, v changes nonlinearly (because v = uf/(u - f)). Differentiating shows the image velocity depends on u and hence varies with time. Initially, when object is far, image moves slowly; as the object comes nearer, image speed increases and can even tend to infinity as the object approaches the focal point. Thus the image moves away with non-uniform acceleration.

Q.4. A passenger in an aeroplane shall
(a) Never see a rainbow.
(b) May see a primary and a secondary rainbow as concentric circles.
(c) May see a primary and a secondary rainbow as concentric arcs.
(d) Shall never see a secondary rainbow.

Correct Answer is Option (b)

Explanation:

From a ground observer a rainbow appears as an arc because the observer's horizon blocks the lower part of the circle. A passenger in an aeroplane above most of the ground obstruction may see the complete circular form of the primary and secondary rainbows; these appear as concentric circles because the secondary rainbow is formed by two internal reflections and subtends a larger angular radius about the antisolar point. Hence option (b) is correct.

Note (related remark):

If an object moves with constant speed V₀ towards a convex lens from infinity to focus, the image moves with non-uniform speed. For an object approaching the lens, the image moves away with non-uniform acceleration; the image speed can be expressed using magnification relations but depends nonlinearly on object position.

Q.5. You are given four sources of light each one providing a light of a single colour - red, blue, green and yellow. Suppose the angle of refraction for a beam of  yellow light corresponding to a particular angle of incidence at the interface of two media is 90°. Which of the following statements is correct if the source of yellow light is replaced with that of other lights without changing the angle of incidence?
(a) The beam of red light would undergo total internal reflection.
(b) The beam of red light would bend towards normal while it gets refracted through the second medium.
(c) The beam of blue light would undergo total internal reflection.
(d) The beam of green light would bend away from the normal as it gets refracted through the second medium.

Correct Answer is Option (c)

Explanation:

If, for a given angle of incidence, the refracted ray for yellow light emerges at 90°, this incidence angle is the critical angle for yellow light. The light is going from a denser to a rarer medium. The critical angle c satisfies sin c = 1/μ (with μ the refractive index for that colour in the denser medium). Dispersion implies refractive index depends on wavelength: typically μ(violet) > μ(blue) > μ(green) > μ(yellow) > μ(red). Thus the critical angles follow the reverse order: c(violet) < c(blue) < c(green) < c(yellow)

Given that the chosen incidence angle equals c(yellow), any colour with smaller critical angle (blue, violet, green if their critical angles are smaller than c(yellow)) will have incidence > their critical angle and therefore undergo total internal reflection. Red, having the largest critical angle, will not TIR for this incidence angle and will refract into the rarer medium. Among the choices, blue will certainly undergo total internal reflection; hence option (c) is correct.

Q.6. The radius of curvature of the curved surface of a plano-convex lens is 20 cm. If the refractive index of the material of the lens be 1.5, it will
(a) Act as a convex lens only for the objects that lie on its curved side.
(b) Act as a concave lens for the objects that lie on its curved side.
(c) Act as a convex lens irrespective of the side on which the object lies.
(d) Act as a concave lens irrespective of side on which object lies.

Correct Answer is Option (c)

Explanation and steps:

Use the lens maker's formula for a thin lens in air:

1/f = (μ - 1)(1/R₁ - 1/R₂).

Case 1: Object on the curved side (curved surface first): R₁ = +20 cm, R₂ = ∞, μ = 1.5, μ₁(air)=1.

1/f = (1.5 - 1)(1/20 - 0) = 0.5 × 1/20 = 1/40 ⇒ f = +40 cm.

Case 2: Object on the plane side (plane surface first): R₁ = ∞, R₂ = -20 cm (sign convention),

1/f = (1.5 - 1)(0 - (-1/20)) = 0.5 × (1/20) = 1/40 ⇒ f = +40 cm.

In both orientations the focal length is +40 cm, so the lens acts as a converging (convex) lens irrespective of the side on which the object lies. Thus option (c) is correct.

Q.7. The phenomena involved in the reflection of radio waves by the ionosphere is similar to
(a) Reflection of light by a plane mirror.
(b) Total internal reflection of light in the air during a mirage.
(c) Dispersion of light by water molecules during the formation of a rainbow.
(d) Scattering of light by the particles of air.

Correct Answer is Option (b)

Explanation:

The ionosphere has a refractive index that varies with electron density and the radio waves can be turned back towards Earth when propagating from a region of higher effective refractive index to lower-this is analogous to total internal reflection in optics where a wave in a denser medium is reflected at a boundary with a rarer medium when the incidence angle exceeds the critical angle. Hence the ionospheric reflection of radio waves is most like total internal reflection (option b).

Q.8. The direction of ray of light incident on a concave mirror is shown by PQ while directions in which the ray would travel after reflection is shown by four rays marked 1, 2, 3 and 4 (Fig.). Which of the four rays correctly shows the direction of reflected ray?

(a) 1
(b) 2
(c) 3
(d) 4

Correct Answer is Option (b)

Reasoning:

The incident ray PQ passes through the focus F before striking the concave mirror. A ray that passes through the focus is reflected parallel to the principal axis. Ray marked 2 is parallel to the principal axis after reflection, so option (b) is correct.

Important points: (rays used for locating images of extended objects)

• A ray initially parallel to the principal axis is reflected through the focus of the mirror.
• A ray passing through the centre of curvature is reflected back along itself.
• A ray initially passing through the focus is reflected parallel to the principal axis.
• A ray incident at the pole is reflected symmetrically (angle of incidence = angle of reflection measured from the normal at the pole).

Q.9. The optical density of turpentine is higher than that of water while its mass density is lower. Fig. shows a layer of turpentine floating over water in a container. For which one of the four rays incident on turpentine in Fig, the path shown is correct?

 (a) 1
(b) 2
(c) 3
(d) 4

Correct Answer is Option (b)

Explanation:

Given optical densities: μ_air < μ_turpentine="" /> μ_water. A ray incident from air enters turpentine (rarer → denser) and bends towards the normal. On passing from turpentine into water (denser → less dense if water is optically less dense than turpentine?-here μ_turpentine > μ_water), the ray bends away from the normal. Thus the path that first bends towards the normal and then away (as shown for ray 2) is correct.

Q.10. A car is moving with at a constant speed of 60 km h^-1 on a straight road. Looking at the rear view mirror, the driver finds that the car following him is at a distance of 100 m and is approaching with a speed of 5 km h^-1. In order to keep track of the car in the rear, the driver begins to glance alternatively at the rear and side mirror of his car after every 2 s till the other car overtakes. If the two cars were maintaining their speeds, which of the following statement (s) is/are correct?
(a) The speed of the car in the rear is 65 km h^-1.
(b) In the side mirror the car in the rear would appear to approach with a speed of 5 km h^-1 to the driver of the leading car.
(c) In the rear view mirror the speed of the approaching car would appear to decrease as the distance between the cars decreases.
(d) In the side mirror, the speed of the approaching car would appear to increase as the distance between the cars decreases.

Correct Answer is Option (d)

Explanation and remarks:

(a) If the leading car moves at 60 km/h and the relative approach speed is 5 km/h, the speed of the rear car is 60 + 5 = 65 km/h. So statement (a) is true (but the official correct selection given here emphasises optical appearance rather than raw speeds).

Plane mirrors produce virtual, erect images whose apparent size depends on object distance from the mirror. For a driver looking in the side mirror (which typically provides a small field and reduced image), the apparent angular approach speed can change because magnification of a plane mirror image depends on geometry. In the rear view mirror (usually a plane mirror), as the distance decreases the image appears to approach faster (apparent speed increases) rather than decrease. Therefore statement (c) is not correct.

In many side mirrors designed to give a wide field, the apparent speed of an approaching car may appear to increase as distance decreases due to changing angular size and perspective; this supports statement (d). The official chosen option emphasises the appearance in the side mirror, hence option (d) is marked correct here.

Optical note: For objects along the principal axis of a mirror, differentiating the mirror formula shows the image velocity depends on object position; generally the image motion is nonlinearly related to object motion.

Q.11. There are certain material developed in laboratories which have a negative refractive index (Fig.). A ray incident from air (medium 1) into such a medium (medium 2) shall follow a path given by
(a) 

(b) 

(c) 

(d) 

Correct Answer is Option (a)

Explanation:

Metamaterials with negative refractive index reverse the usual direction of refraction predicted by Snell's law. For an incident ray from a positive-index medium (air) into a negative-index medium the refracted ray appears on the same side of the normal as the incident ray (effectively bending 'the other way'). The ray configuration shown in option (a) corresponds to this behaviour.

Q.12. Consider an extended object immersed in water contained in a plane trough. When seen from close to the edge of the trough the object looks distorted because
(a) The apparent depth of the points close to the edge are nearer the surface of the water compared to the points away from the edge.
(b) The angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air.
(c) Some of the points of the object far away from the edge may not be visible because of total internal reflection.
(d) Water in a trough acts as a lens and magnifies the object.

Correct Answer is Option (a), (b) & (c)

Explanation:

• Apparent depth variation: The apparent upward shift of an immersed object depends on the local depth; when viewed from the edge the relative differences make the nearer end appear less shifted than the farther end, producing a distortion. This verifies (a).
• Angular subtension: Because the image is formed closer to the surface (apparent depth less than real depth), the angle subtended at the eye by the image can be different (often smaller) than the angle subtended in air, supporting (b).
• Total internal reflection: Rays from far points of the object may meet the water-air interface at angles exceeding the critical angle and will be totally internally reflected, so those points may not be visible. This verifies (c).

Option (d) is not the principal reason for the distortion in the described situation: the trough geometry and refraction at the surface produce the observed effects rather than the trough acting as a lens in the usual focussed-imaging sense.

Q.13. A rectangular block of glass ABCD has a refractive index 1.6. A pin is placed midway on the face AB (Fig.). When observed from the face AD, the pin shall

(a) Appear to be near A.
(b) Appear to be near D.
(c) Appear to be at the centre of AD.
(d) Not be seen at all.

Correct Answer is Option (a) & (b)

Explanation:

Let point P be at the midpoint of face AB. Rays from P travelling towards the face AD will emerge at different points on AD. Near point A (on AD), the incidence angles at the internal surface are smaller than the critical angle and refraction allows rays to emerge; such rays form an apparent image P' nearer to both A and D than P. Near point D, for rays making larger incidence angles the angle of incidence may exceed the critical angle (i_c ≈ 38.7° for μ = 1.6), producing total internal reflection and preventing visibility through some parts of AD. Hence P may appear shifted nearer A and also nearer D when seen through parts of AD where rays can exit. Therefore options (a) and (b) are both supported.

Q.14. Between the primary and secondary rainbows, there is a dark band known as Alexander's dark band. This is because
(a) Light scattered into this region interfere destructively.
(b) There is no light scattered into this region.
(c) Light is absorbed in this region. 
(d) Angle made at the eye by the scattered rays with respect to the incident light of the sun lies between approximately 42° and 50°.

Correct Answer is Option (a) & (d)

Explanation:

The primary rainbow is produced by one internal reflection in raindrops and is observed at scattering angles near 42°. The secondary rainbow, produced by two internal reflections, is observed at larger scattering angles near 51°-54°. Between these two angular ranges (approximately 42°-50°) there is a relative deficit of scattered intensity leading to the dark region called Alexander's dark band. Interference effects of scattered rays and the geometry of scattering cause the observed diminished intensity in this angular range. Thus (a) and (d) are correct.

Q.15. A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in
(a) A larger angle to be subtended by the object at the eye and hence viewed in greater detail.
(b) The formation of a virtual erect image.
(c) Increase in the field of view.
(d) Infinite magnification at the near point.

Correct Answer is Option (a) & (b)

Explanation:

A magnifying glass (simple convex lens) forms a virtual, erect and enlarged image when the object is placed within the focal length. Because the object can be brought closer than the unaided near point, the object subtends a larger angle at the eye and so finer details are visible-this is the working principle of a magnifier. Option (c) is not generally correct (field of view depends on lens design) and (d) is not correct as magnification is finite and determined by lens focal length and eye accommodation.

Q.16. An astronomical refractive telescope has an objective of focal length 20m and an eyepiece of focal length 2cm.
(a) The length of the telescope tube is 20.02m.
(b) The magnification is 1000.
(c) The image formed is inverted.
(d) An objective of a larger aperture will increase the brightness and reduce chromatic aberration of the image.

Correct Answer is Option (a), (b) & (c)

Explanation and steps:

Objective focal length f_o = 20 m; eyepiece focal length f_e = 2 cm = 0.02 m.

Length of the telescope (normal astronomical arrangement with final image at infinity) L = f_o + f_e = 20 + 0.02 = 20.02 m. Thus (a) is true.

Angular magnification m = f_o / f_e = 20 / 0.02 = 1000. Thus (b) is true.

In a refracting astronomical telescope the image produced by the objective is real and inverted; the eyepiece produces a magnified virtual image of this real image, so the final viewed image is inverted. Thus (c) is true.

Regarding (d): a larger objective aperture increases light-gathering power and hence brightness, and improves angular resolution (reduces diffraction). However, chromatic aberration is mainly determined by lens material dispersion and focal length design; simply increasing aperture does not reduce chromatic aberration (it can make some aberrations more noticeable unless corrected by achromatic combinations). Hence (d) is not generally correct.

Very Short Answer Type Questions

Q.17. Will the focal length of a lens for red light be more, same or less than that for blue light?

Answer and explanation:

The refractive index μ depends on wavelength; Cauchy's empirical form is μ = A + B/λ² + C/λ⁴ + ···. Since λ_red > λ_blue, μ_red < μ_blue.="" from="" the="" lens="" maker's="" />

1/f = (μ - 1)(1/R₁ - 1/R₂),

smaller μ gives smaller (μ - 1) and therefore larger focal length. Thus f_red > f_blue; the focal length for red light is greater than that for blue light.

Q.18. The near vision of an average person is 25cm. To view an object with an angular magnification of 10, what should be the power of the microscope?

Solution:

For a simple magnifier used with comfortable near point D = 25 cm and angular magnification m = D/f.

Therefore f = D/m = 25 cm / 10 = 2.5 cm = 0.025 m.

Power P = 1/f (in metres) = 1/0.025 = 40 dioptres (D).

Hence the required power is 40 D.

Q.19. An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the lens is reversed?

Answer and explanation:

The thin lens formula 1/v - 1/u = 1/f shows that for a fixed object position and a fixed focal length f the image position v is determined only by f and u. From the lens maker's formula

1/f = (μ - 1)(1/R₁ - 1/R₂).

Reversing the lens changes the sign convention of R₁ and R₂ but the combination (1/R₁ - 1/R₂) becomes (1/R₂ - 1/R₁) with appropriate signs; for a symmetric switching the numerical value of f remains the same. Therefore the image position does not change on reversing the lens (assuming the lens is thin and in air).

Q.20. Three immiscible liquids of densities d₁ > d₂ > d₃ and refractive indices µ₁ > µ₂ > µ₃ are put in a beaker. The height of each liquid column is h/3. A dot is made at the bottom of the beaker. For near normal vision, find the apparent depth of the dot.

Solution:

Arrange liquids from bottom to top as layers of refractive index μ₁ (bottom), μ₂ (middle) and μ₃ (top), each of real thickness h/3. For near-normal viewing, the apparent depth contributed by each layer is its real thickness divided by the refractive index of the layer through which the final rays emerge to the observer. Tracing successive refractions, the net apparent depth x₃ (measured from the top surface) is

x₃ = h/3 (1/μ₃ + 1/μ₂ + 1/μ₁).

Thus the apparent depth is the sum of the contributions h/3 divided by the corresponding refractive indices in the order the light traverses outward.

Q.21. For a glass prism (µ = √3) the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism.

Solution:

For minimum deviation in a prism the relation μ = sin[(A + δ_m)/2] / sin(A/2) simplifies for symmetric passage to (μ) = sin[(A + δ_m)/2] / sin(A/2). Given δ_m = A, substitute to get:

μ = sin(A) / sin(A/2) = 2 cos(A/2).

Given μ = √3, so 2 cos(A/2) = √3 ⇒ cos(A/2) = √3/2 ⇒ A/2 = 30° ⇒ A = 60°.

Hence the prism angle is 60°.

Short Answer Type Questions

Q.22. A short object of length L is placed along the principal axis of a concave mirror away from focus. The object distance is u. If the mirror has a focal length f, what will be the length of the image? You may take L << |v - f|.

Solution and steps:

Consider the object whose two ends are at u₁ = u - L/2 and u₂ = u + L/2. Let images of these ends form at v₁ and v₂ respectively. Image length L' = |v₁ - v₂|.

Use mirror formula for each end: 1/v + 1/u = 1/f so v = uf/(u - f).

For the two close object positions expand v to first order in L (treat L small): the differential approach gives

d v/du = -fu/(u - f)².

Hence approximate change in image position for object end separation L is

L' ≈ |dv/du| × L = [f²/(u - f)²] × L.

So L' = L f²/(u - f)² (valid when L ≪ |u - f|).

Q.23.  A circular disc of radius 'R ' is placed co-axially and horizontally inside an opaque hemispherical bowl of radius 'a' (Fig.). The far edge of the disc is just visible when viewed from the edge of the bowl. The bowl is filled with transparent liquid of refractive index µ and the near edge of the disc becomes just visible. How far below the top of the bowl is the disc placed?

Solution outline and steps:

Consider rays from the far and near edges of the disc reaching the observer at the rim of the hemisphere. Using geometry of the hemisphere and Snell's law at the liquid-air surface one relates the geometry (distances OM, CN, BN etc.) to the critical/refraction angles. The condition for visibility at the rim gives a relation between the disc depth below the top and the parameters a, R and μ. Following the geometric construction and applying Snell's law at the exit point M:

sin i / sin r = 1/μ (air on outside), where r is determined from triangle geometry at M. Solving these relations yields the required depth expression (see figure relations in the diagram for complete algebraic steps).

Q.24. A thin convex lens of focal length 25 cm is cut into two pieces 0.5 cm above the principal axis. The top part is placed at (0, 0) and an object placed at (-50 cm, 0). Find the coordinates of the image.

Solution:

Object distance u = -50 cm. Focal length f = +25 cm. Using lens formula 1/v - 1/u = 1/f,

1/v - 1/(-50) = 1/25 ⇒ 1/v + 1/50 = 1/25 ⇒ 1/v = 1/50 ⇒ v = +50 cm.

Linear magnification m = v/u = 50/(-50) = -1 ⇒ image size equals object size and is inverted. The object was 0.5 cm above the axis; inversion places the image 0.5 cm below the axis.

Hence the image coordinates are (x, y) = (50 cm, -0.5 cm).

Q.25. In many experimental set-ups the source and screen are fixed at a distance say D and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen.
Find the distance between these points and the ratio of the image sizes for these two points.

Solution and steps (displacement method):

Place a convex lens between an object O and a screen S separated by distance D. Let two lens positions I and II give sharp images on the screen. Let the lens at position I be at distance x₁ from the object; then image distance v₁ = D - x₁. At position II the lens is at x₂ from the object, with v₂ = D - x₂.

Apply lens formula for both positions and compare. One finds the two nontrivial solutions satisfy x₂ = D - x₁. Thus if the first lens position is at x₁ from the object the second is at D - x₁, so the distance between the two positions is

d = |x₂ - x₁| = |D - 2x₁|.

Expressing d in terms of D and f by eliminating x₁ using the lens equation leads to

d² = D² - 4Df.

Magnifications for the two positions are m₁ = (D - x₁)/x₁ and m₂ = (D - x₂)/x₂. Using x₂ = D - x₁ gives

m₁ = (D - d)/(D + d), m₂ = (D + d)/(D - d),

and thus m₂/m₁ = [(D + d)/(D - d)]².

Important points:

• A solution for d exists only when D ≥ 4f.
• When D = 4f, there is exactly one position (the two coincide) where a sharp image forms.
• When D > 4f there are two positions producing sharp images on the fixed screen; this is the basis of the displacement method to determine focal length f.

Q.26. A jar of height h is filled with a transparent liquid of refractive index µ (Fig.). At the centre of the jar on the bottom surface is a dot. Find the minimum diameter of a disc, such that when placed on the top surface symmetrically about the centre, the dot is invisible.

Solution outline:

The dot becomes invisible when all rays from the dot that could emerge from the liquid to the air are blocked. The marginal ray that just grazes the top surface corresponds to incidence at the critical angle i_c, where sin i_c = 1/μ. From geometry, the lateral half-width of the disc P is related to the jar height h and the critical angle by

tan i_c = (d/2)/h ⇒ d_min = 2h tan i_c = 2h √(1 - 1/μ²)/ (1/μ) = 2h √(μ² - 1)/1.

Thus, using sin i_c = 1/μ, one can obtain the minimum disc diameter (the figure steps show the geometric derivation).

Q.27. A myopic adult has a far point at 0.1 m. His power of accommodation is 4 diopters.
(i) What power lenses are required to see distant objects?
(ii) What is his near point without glasses?
(iii) What is his near point with glasses? (Take the image distance from the lens of the eye to the retina to be 2 cm.)

Solution and steps:

(i) For a myopic eye whose far point is at 0.1 m, the eye in relaxed state focuses parallel rays at the retina only for objects at 0.1 m. To see distant objects (object at ∞), corrective lenses should form a virtual image of an object at ∞ at the person's far point 0.1 m.

Lens power required P_g satisfies 1/∞ + P_g = 1/(-0.1) in sign convention for spectacle lens making object at ∞ seen as at -0.1 m (virtual object), so P_g = -10 D. Thus a concave lens of -10 D is required.

(ii) Power of normal eye at far point P_f = 1/0.1 + 1/0.02 = 60 D (if the relaxed eye lens-power plus eye's internal lens combine to produce focus at retina distance 2 cm). Given accommodation range is 4 D, the near point for this person without glasses corresponds to additional power of 4 D beyond the relaxed power: P_n = P_f + 4 = 64 D ⇒ near point distance x_n = 1/P_n = 1/64 ≈ 0.015625 m ≈ 1.56 cm. (This gives a near point very close because of high total power; in practice such small numbers reflect model assumptions including the 2 cm retina distance.)

(iii) With glasses of power -10 D, the eye plus lens system power changes and the near point is shifted outward; by combining lens powers and using 1/f_total = P_total etc., the new near point can be computed. Using the diagrammatic relations and formulae for combined eye-lens system (detailed algebra shown in the referenced figure), the near point with glasses is obtained (see figure steps for algebraic substitutions).

Long Answer Type Questions

Q.28. Show that for material with refractive index µ ≥ √2 , light incident at any angle shall be guided along a length perpendicular to the incident face.

Solution and steps:

Consider a slab with two faces AB (incident face) and AC (face perpendicular to AB). A ray incident on AB at angle i refracts inside making angle r to the normal and then meets AC. The ray will be guided along the AC face if the angle φ that the ray makes with AC is ≥ critical angle φ_c for total internal reflection at AC.

From geometry φ + r = 90° ⇒ sin φ = cos r. At criticality sin φ_c = cos r_c, and total internal reflection at AC requires sin φ_c ≥ 1/μ ⇒ cos r ≥ 1/μ.

Using Snell's law at AB: sin i = μ sin r. The maximum possible sin i is 1, so the condition for all incident angles to be guided is achieved when even for i = 90° the corresponding cos r ≥ 1/μ.

Setting i = 90°, from sin i = μ sin r ⇒ 1 = μ sin r ⇒ sin r = 1/μ ⇒ cos r = √(1 - 1/μ²). The guiding condition cos r ≥ 1/μ becomes √(1 - 1/μ²) ≥ 1/μ ⇒ 1 - 1/μ² ≥ 1/μ² ⇒ 1 ≥ 2/μ² ⇒ μ² ≥ 2 ⇒ μ ≥ √2.

Therefore if μ ≥ √2, rays incident at any angle on AB will be guided along AC (subject to geometric constraints); this proves the required result.

Q.29. The mixture a pure liquid and a solution in a long vertical column (i.e, horizontal dimensions << vertical dimensions) produces diffusion of solute particles and hence a refractive index gradient along the vertical dimension. A ray of light entering the column at right angles to the vertical is deviated from its original path. Find the deviation in travelling a horizontal distance d << h, the height of the column.

Solution and steps:

Let refractive index vary with height y: μ(y). Consider a small segment of the ray between x and x + dx where the ray makes angle θ(x) with the horizontal and is at height y(x). By Snell's law along this slowly varying medium, μ(y) sin θ = constant along the ray.

Differentiate: μ d(sin θ) + sin θ dμ = 0 ⇒ μ cos θ dθ + sin θ (dμ/dy) dy = 0 ⇒ dθ = -(1/μ)(dμ/dy) tan θ dy.

Relate dy to dx: tan θ = dy/dx ⇒ dy = tan θ dx. Substitute into dθ equation and convert to differential relation in x. Integrate across small horizontal distance d (with appropriate approximations; take k = dμ/dy approximate constant over small h): the net deviation Δθ or transverse displacement can be obtained to first order in k and d. The detailed integration (shown in the figure steps) yields the transverse deviation of the emergent ray from its original straight-line path proportional to k d²/(2μ) for paraxial small-angle approximations.

The full derivation with intermediate integrals is shown in the figure below.

Q.30. If light passes near a massive object, the gravitational interaction causes a bending of the ray. This can be thought of as happening due to a change in the effective refractive index of the medium given by
n(r) = 1 + 2 GM/rc²
where r is the distance of the point of consideration from the centre of the mass of the massive body, G is the universal gravitational constant, M the mass of the body and c the speed of light in vacuum. Considering a spherical object find the deviation of the ray from the original path as it grazes the object.

Solution and steps:

Treat gravitational bending as refraction through a medium with spherically symmetric refractive index n(r) ≈ 1 + 2GM/(rc²). Use Snell's law applied in differential shells n(r) sin θ(r) = constant. Differentiate and integrate radially from large negative x to large positive x across a grazing path at impact parameter R (closest approach equal to the sphere radius for grazing). Using approximations n(r) ≈ 1 + ε(r) with ε small and linearising, integrate the incremental angular changes to find total deflection Δ ≈ 4GM/(Rc²).

This reproduces the classical result from weak-field general relativity (to leading order), showing the deflection angle for a grazing ray: Δ ≈ 4GM/(Rc²).

The diagrammatic algebraic steps with substitutions r² = x² + R² and a change of variables x = R tan φ are shown in the figure steps.

Q.31. An infinitely long cylinder of radius R is made of an unusual exotic material with refractive index -1 (Fig.). The cylinder is placed between two planes whose normals are along the y direction. The center of the cylinder O lies along the y-axis. A narrow laser beam is directed along the y direction from the lower plate. The laser source is at a horizontal distance x from the diameter in the y direction. Find the range of x such that light emitted from the lower plane does not reach the upper plane.

Solution outline and key steps:

Materials with μ = -1 invert the sign of the refracted angle; the negative refractive index leads to negative refraction inside the cylinder. Trace the ray entering the cylinder, refracted twice (entry and exit), and compute the net angular deviation of the emergent ray with respect to the incident beam. For the given geometry the total deviation can be shown to be 4θ_r (where θ_r is the angle inside related by Snell's law with negative index). The emergent beam will miss the upper plate if the lateral displacement at the top exceeds the half-separation between plates.

Solving the trigonometric geometry with Snell's law adapted for μ = -1 gives an inequality for x in terms of R and the plate separation; the algebraic steps and final inequality are shown in the figure steps. The core result is that rays originating beyond a certain lateral distance x_max will be deflected sufficiently to not reach the receiving plate; the exact x-range is obtained from the derived expressions.

Q.32. (i) Consider a thin lens placed between a source (S) and an observer (O) (Fig.). Let the thickness of the lens vary as w(b) = ω₀ - (b²/α), where b is the verticle distance from the pole. ω₀ is a constant. Using Fermat's principle i.e. the time of transit for a ray between the source and observer is an extremum, find the condition that all paraxial rays starting  from the source will converge at a point O on the axis. Find the focal length.

(ii) A gravitational lens may be assumed to have a varying width of the form
w(b) = k₁ln(k₂ / b)     b_min < b < b_max
= k₁ln(k₂ / b_min)   b < b_min
Show that an observer will see an image of a point object as a ring about the center of the lens with an angular radius 

Solution - Part (i):

Compute optical path time t(b) from S to O for a paraxial ray passing at height b. Expand geometrical path lengths using paraxial approximation and include extra time inside the lens where refractive index n gives time (n - 1)w(b)/c. The total time:

t(b) = (u + v)/c + (b²/2c)(1/u + 1/v) + (n - 1)w(b)/c.

Fermat's principle requires dt/db = 0 for all small b (paraxial rays). Differentiating and setting coefficient of b to zero gives the condition:

1/D - 2(n - 1)/α = 0 where 1/D = 1/u + 1/v.

Thus α = 2(n - 1)D. With this choice all paraxial rays from S converge to O and D acts as the focal parameter; for the symmetric case with object at infinity (u → ∞) this yields the focal length f = D = v.

Solution - Part (ii):

For the gravitational-lens analogue with w(b) = k₁ ln(k₂/b) the time difference for rays at different b leads to stationarity conditions which relate b to the impact parameter and produce constructive interference over a ring of certain angular radius. Differentiation of t(b) with respect to b and setting to zero selects those b that contribute to the image; converting to an angular variable gives the observed ring radius (the Einstein ring) with angular radius proportional to √(4GM/(c²D_eff)) in astrophysical notation. The algebra is presented in the figure steps and leads to the expression shown in the figure placeholder.