NEET Ray Optics and Optical Instruments Previous Year Questions with Solutions

The slab of medium Q has a thickness of 5 cm. Since object O is located at its center:

• • d_1 is the real depth from the upper P-Q interface.
• • d_2 is the real depth from the lower Q-R interface.

Observer is in medium P (n_P = 1), looking at object O in medium Q (n_Q = 1.25):

• • The object appears shifted closer to the interface by 0.5 cm.

Observer is in medium R (n_R = 1.5), looking at object O in medium Q (n_Q = 1.25):

• • Because n_R > n_Q, the object appears further from the interface.

Subtract the two apparent depths:

• • Matches Option (2) exactly.

For the convex lens L_1, focal length f_1 = +10 cm and object distance u_1 = -30 cm:

• • v_1 = +15 cm.
• • The real image I_1 is formed 15 cm to the right of lens L_1.

The second lens L_2 is located 3 cm to the right of L_1. The image I_1 acts as a virtual object for L_2:

• • Because the rays are converging toward a point 12 cm behind L_2, object distance u_2 is positive: u_2 = +12 cm.

For the diverging concave lens L_2, focal length is f_2 = -10 cm:

• • v_2 = -60 cm.
• • The negative sign indicates the final image is formed 60 cm to the left of the concave lens L_2.
• • Matches Option (2) exactly.

For a compound microscope with tube length L much larger than the focal lengths of the lenses, the total magnifying power m for an image formed at infinity (normal adjustment, minimum eye strain) is the product of the magnifications of objective and eyepiece.

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

Substitute the given optical parameters into the formula:

• • Matches Option (2) with 100% precision.

For n thin lenses placed axially in contact, the equivalent focal length F satisfies 1/F = 1/f₁ + 1/f₂ + ... + 1/f_n. Since power P = 1/f:

• • Power is strictly additive for lenses in contact.

The image formed by the first lens acts as the object for the second lens, and so forth. Consequently, the total lateral magnification produced by the lens system is the product of the magnifications produced by each individual lens:

• • Thus, the combination has power 4p and magnification m⁴, corresponding to Option (3).

Light enters the prism from air (n = 1) at point P on face AB with angle of incidence i = 30°. By Snell’s law:

The ray emerges along the face AC, which means the angle of refraction in air is 90°. Therefore, the internal angle of incidence r2 at face AC must be identically equal to the critical angle θ_c for the prism-air interface:

For any triangular prism of refracting angle A, the internal angles satisfy r1 + r2 = A. The figure indicates that vertex A is a right angle (A = 90°):

Substituting r2 = 90° - r1 into the critical angle formula:

Using the fundamental trigonometric identity sin² r1 + cos² r1 = 1:

Inverting and taking the positive square root:

For an astronomical refracting telescope viewing a distant celestial object under normal adjustment (relaxed eye, final image at infinity), the angular magnification m is given by the ratio of the focal length of the objective lens to that of the eyepiece lens:

Substituting f_o = 140 cm and f_e = 5.0 cm:

When light travels from an optically denser medium of refractive index μ into an optically rarer medium (air, μ_air ≈ 1), total internal reflection occurs at the critical angle of incidence θ_c where the angle of refraction is 90°:

By the wave definition of refractive index, μ is the ratio of the speed of light in air (v_1) to the speed of light in the denser medium (v_2):

Using the kinematic relation speed = distance / time for each medium:

Substitute v_1 and v_2 into the critical angle relation:

The given optical system consists of three thin lenses placed in contact along a common optical axis: (i) a first plano-concave lens made of the medium of refractive index n₂ = 1.6, (ii) an equiconvex lens made of material of refractive index n₁ = 1.5, and (iii) a second plano-concave lens made of the medium of refractive index n₂ = 1.6.

Using the Lens Maker’s Formula for the central equiconvex lens in air with R₁ = +20 cm and R₂ = -20 cm:

For the left plano-concave lens, the first surface is plane (R = ∞) and the second surface has radius of curvature R = +20 cm:

Because the right plano-concave lens is identical in curvature and material, 1/f₃ = -3/100 cm⁻¹. Summing all three powers:

Inverting yields the net focal length of the combination:

When two thin lenses of focal lengths f₁ and f₂ are placed coaxially in direct physical contact, their powers add algebraically:

By optical sign convention, a convex (converging) lens has a positive focal length, while a concave (diverging) lens of identical magnitude has a negative focal length:

Substitute the focal lengths into the power equation:

Taking the reciprocal of zero power yields:

Astronomical telescopes observe extremely distant, faint celestial bodies. The amount of light gathered from a distant star is directly proportional to the surface area of the objective lens:

Due to diffraction at the circular aperture of diameter $D$, the angular limit of resolution is $\Delta\theta = \frac{1.22\lambda}{D}$. The resolving power ($RP$) is the reciprocal of $\Delta\theta$:

Because a larger aperture collects more light photons and reduces diffraction blurring, the resultant image has superior brightness, sharpness, and visibility. Hence, all statements (1), (2), and (3) are scientifically true.

The incident ray is perpendicular to the hypotenuse face (marked with a right-angle symbol in the diagram). By Snell's law, $\sin i_1 = 0 \implies \sin r_1 = 0$, so the ray enters the prism without any deviation.

Consider the triangle at the top of the prism: - The apex angle between the vertical face and the hypotenuse is $A = 180^\circ - (90^\circ + 60^\circ) = 30^\circ$. - The ray enters perpendicular to the hypotenuse, making an angle of $90^\circ$ with the hypotenuse. - Inside the top small triangle, the angle between the ray and the vertical face is $180^\circ - (90^\circ + 30^\circ) = 60^\circ$. - Since the normal to the vertical face is perpendicular to it, the angle of incidence $r_2$ (angle with the normal) is:

At the vertical glass-air interface, light travels from glass ($\mu = \sqrt{3}$) into air ($n = 1$):

A parallel beam of light falling on convex lens A has object distance $u_1 = \infty$. By the lens formula, the image is formed at its principal focus:

This real converging image acts as a virtual object for concave lens B placed at distance $d$ from A. The object distance for lens B is $u_2 = +(20 - d)$. For the emergent beam to be parallel, the final image must be formed at infinity ($v_2 = \infty$):

Equating $u_2 = 20 - d$ to $5\text{ cm}$:

For a convex lens of focal length $f = +30\text{ cm}$ with object placed at $u = -60\text{ cm}$:

The plane mirror is placed at a distance of $40\text{ cm}$ behind the lens. The converging rays from the lens are intercepted before they reach $v = +60\text{ cm}$. Hence, the point $I_1$ at $60\text{ cm}$ acts as a virtual object for the plane mirror:

A plane mirror reflects light symmetrically. The reflected rays converge to form an image at a distance of $20\text{ cm}$ from the mirror surface. In the official NTA examination key, this is classified as Option (4):

The ray emerges normally from the second surface of the prism. This means the ray is perpendicular to the second interface ($$\theta = 90^\circ$$ to the surface, hence along the normal): $$r_2 = 0^\circ \quad \text{and} \quad e = 0^\circ$$

For any triangular prism, geometry dictates that the prism angle $$A$$ equals the sum of the two internal refraction angles: $$r_1 + r_2 = A$$ Since $$r_2 = 0$$: $$r_1 + 0 = A \implies r_1 = A$$

At the first refracting surface, light passes from air ($$\mu_1 = 1$$) into the prism medium ($$\mu_2 = \mu$$) at incident angle $$i$$: $$\sin i = \mu \sin r_1$$ Substituting $$r_1 = A$$: $$\sin i = \mu \sin A$$

Because the question explicitly specifies a "small angle prism" (typically $$A \le 10^\circ$$), we utilize the standard paraxial small-angle trigonometric approximations: $$\sin i \approx i \quad \text{and} \quad \sin A \approx A$$ Substituting these approximations yields: $$i \approx \mu A$$ This directly gives Option (3).