Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
One side of an equilateral prism is painted by a transparent material of refractive index $n_2$. The refractive index of prism is 1.6. The minimum value of $n_2$ required for total internal reflection from painted face is ______. [Figure: An equilateral prism with its bottom face painted with material of refractive index $n_2$, with a light ray entering from the left side]
(1) $\frac{3\sqrt{3}}{1.6}$
(2) $\sqrt{3}$
(3) $\frac{3.2}{\sqrt{3}}$
(4) $\frac{4\sqrt{3}}{5}$
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Answer: Option (4)
Key idea: Total Internal Reflection in Prisms
For an equilateral prism, all angles are $60°$. When light enters one face and hits the painted base, use geometry: $r_1 + r_2 = A = 60°$
For total internal reflection at the painted face: $\sin(\theta_c) = \frac{n_2}{n_1}$ where $n_1 = 1.6$ is the prism's refractive index
To find minimum $n_2$, maximize the angle of incidence $r_2$ at the painted face. This occurs when $r_1 = 0°$ (perpendicular entry)
Therefore, maximum angle at painted face: $r_2 = 60°$
If sunlight is focused on a paper using convex lens, it starts burning the paper in shortest time when the lens is kept at 30 cm above the paper. If the radius of curvature of the lens is 60 cm then the refractive index of the lens material is $\frac{\alpha}{10}$. The value of $\alpha$ is _______.
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
A rod of length 10 cm lies along the principle axis of a concave mirror of focal length 10 cm as shown in figure. The length of the image is ______ cm.
[Figure: A diagram showing a concave mirror on the right side with a hatched rod of length 10 cm on the left side, positioned 20 cm from the mirror along the principal axis]
(1) 2.5
(2) 5
(3) 7.5
(4) 7
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Answer: Option (2)
Key idea: Image formation by concave mirror for extended object along principal axis
Identify the rod extends from $u_1 = -20$ cm (near end) to $u_2 = -30$ cm (far end) from the concave mirror with $f = -10$ cm
Apply mirror formula for near end: $\frac{1}{v_1} + \frac{1}{u_1} = \frac{1}{f}$ gives $\frac{1}{v_1} = \frac{1}{-10} - \frac{1}{-20} = -\frac{1}{20}$, so $v_1 = -20$ cm
Apply mirror formula for far end: $\frac{1}{v_2} + \frac{1}{u_2} = \frac{1}{f}$ gives $\frac{1}{v_2} = \frac{1}{-10} - \frac{1}{-30} = -\frac{1}{15}$, so $v_2 = -15$ cm
A telescope with objective diameter $R$ is used to observe a distant star emitting light of wavelength 500 nm, at a resolution of $5 \times 10^{-7}$ radian. The value of $R$ is _______ cm.
(1) 61
(2) 122
(3) 244
(4) 305
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Answer: Option (2)
Key idea: Rayleigh Criterion for Angular Resolution of Telescope
Write the Rayleigh criterion formula for angular resolution: $\theta = \frac{1.22\lambda}{D}$
A convex lens is made from glass material having refractive index of 1.4 with same radius of curvature on both sides. The ratio of its focal length and radius of curvature is ________.
(1) $0.5$
(2) $2.5$
(3) $0.8$
(4) $1.25$
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Answer: Option (4)
Key idea: Lensmaker's Equation for Biconvex Lens
Given: Refractive index $n = 1.4$, and equal radii of curvature on both sides
For a biconvex lens: $R_1 = +R$ and $R_2 = -R$ (using sign convention)
A ray of light passing through an equilateral prism is having velocity $2.12 \times 10^8$ m/s in the prism material, then the minimum angle of deviation is ____ degrees.
(1) $45$
(2) $30$
(3) $28$
(4) $58$
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Answer: Option (2)
Key idea: Minimum Deviation in Prism using Refractive Index
A compound microscope is designed with two symmetric biconvex lenses. The objective lens is cut vertically, creating two identical plano-convex lenses. One of them is used in place of original objective lens. To retain same magnification keeping the object distance unchanged, the tube length has to be
(1) increased two times
(2)
(3)
(4)
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Answer: Option (1)
Key idea: Lens Maker's Formula and Compound Microscope Magnification
When a symmetric biconvex lens is cut vertically, each half becomes a plano-convex lens
Using lens maker's formula: For biconvex lens $\frac{1}{f_0} = (\mu-1)\frac{2}{R}$, for plano-convex $\frac{1}{f_1} = (\mu-1)\frac{1}{R}$
Therefore, focal length of plano-convex lens: $f_1 = 2f_0$ (doubles)
For compound microscope, magnification $M = -\frac{L}{f_0} \times \frac{D}{f_e}$ where $L$ is tube length
To maintain same magnification with $f_1 = 2f_0$: $\frac{L_1}{f_0} = \frac{L_2}{2f_0}$
Solving: $L_2 = 2L_1$ (tube length must be doubled)
A thin convex lens and a thin concave lens are kept in contact and are co-axial. Which of the following statements is correct for this combination of two lenses ?
(1) behaves as concave lens if $|f_{convex}| > |f_{concave}|$
(2) behaves as concave lens if $|f_{convex}| < |f_{concave}|$
(3) behaves as convex lens if $|f_{convex}| > |f_{concave}|$
(4) Focal length of the lens system will change if the positions of two lenses are interchanged
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Answer: Option (1)
Key idea: Combination of thin lenses in contact - Power and focal length addition
For two thin lenses in contact: $\frac{1}{f_{combined}} = \frac{1}{f_{convex}} + \frac{1}{f_{concave}}$
Let $f_{convex} > 0$ (positive) and $f_{concave} < 0$ (negative), so $\frac{1}{f_{combined}} = \frac{1}{f_{convex}} - \frac{1}{|f_{concave}|}$
An object $AB$ is placed 15 cm on the left of a convex lens $P$ of focal length 10 cm. Another convex lens $Q$ is now placed 15 cm right of lens $P$. If the focal length of lens $Q$ is 15 cm, the final image is ________.
(1) virtual, formed at 7.5 cm right of lens $Q$, with a size bigger than that of $AB$
(2) real, formed at 7.5 cm right of lens $Q$, with a size same as that of $AB$
(3) formed at infinity.
(4) real, formed at 7 cm right of lens $Q$, with a size smaller than that of $AB$
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Answer: Option (2)
Key idea: Two lens system - successive image formation using lens formula
For lens P: Use lens formula with $u_P = -15$ cm, $f_P = 10$ cm to get $v_P = 30$ cm (real image 30 cm right of P)
A spherical interface lens of radius $R$ separates two media of refractive indices 1 and 1.4 respectively as shown in the figure below. A point source is placed at a distance of $4R$ in front of spherical interface. The magnitude of the magnification of point source image is _______. [Figure: A diagram showing a point source P at distance 4R from a spherical interface of radius R, with refractive indices $n_1 = 1$ on the left and $n_2 = 1.4$ on the right]
(1) $1.66$
(2) $2.33$
(3) $2.66$
(4) $1.33$
Show answer and solution
Answer: Option (1)
Key idea: Refraction at spherical surface and magnification
A concave mirror of focal length 10 cm forms an image which is double the size of object when the object is placed at two different positions. The distance between the two positions of the object is __________ cm.
Show answer and solution
Answer: 10
Key idea: Mirror equation and magnification for concave mirrors with two object positions giving same magnitude of magnification
Given: Concave mirror with $f = -10$ cm, magnification $|m| = 2$
**Case 1:** Real image with $m = -2$, so $v = 2u$. Using $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$: $\frac{-1}{10} = \frac{1}{2u} + \frac{1}{u} = \frac{3}{2u}$, giving $u_1 = -15$ cm
**Case 2:** Virtual image with $m = +2$, so $v = -2u$. Using mirror equation: $\frac{-1}{10} = \frac{-1}{2u} + \frac{1}{u} = \frac{1}{2u}$, giving $u_2 = -5$ cm
Verify: For $u_1 = -15$ cm, $v = -30$ cm, $m = -2$ ✓; For $u_2 = -5$ cm, $v = +10$ cm, $m = +2$ ✓
Distance between two positions = $|u_1 - u_2| = |-15 - (-5)| = 10$ cm
A thin biconvex lens is prepared from the glass ($\mu = 1.5$) both curved surfaces of which have equal radii of 20 cm each. Left side surface of the lens is silvered from outside to make it reflecting. To have the position of image and object at the same place, the object should be placed, from the lens at a distance of _________ cm.
(1) 10
(2) 12.5
(3) 13
(4) 13.5
Show answer and solution
Answer: Option (1)
Key idea: Silvered Lens - Equivalent Focal Length and Center of Curvature
Find focal length of biconvex lens: $\frac{1}{f_L} = (\mu-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 0.5\left(\frac{1}{20} + \frac{1}{20}\right)$, so $f_L = 20$ cm
Silvered left surface acts as concave mirror: $f_m = \frac{R}{2} = \frac{20}{2} = 10$ cm
Light ray incident along a vector $\overrightarrow{AO}$ ($\overrightarrow{AO} = 2\hat{i} - 3\hat{j}$) emerges out along vector $\overrightarrow{OB}$ ($\overrightarrow{OB} = C\hat{i} - 4\hat{j}$) as shown in the figure below. The value of C is _______. [Figure: Light ray passes from medium with $\mu_1 = 1$ through point O to medium with $\mu_2 = 1.5$, with points A, O, and B marked]
(1) 1.6
(2) 0.16
(3) 11.6
(4) 16
Show answer and solution
Answer: Option (1)
Key idea: Snell's Law in Vector Form
Find unit vectors: $\hat{i}_1 = \frac{2\hat{i} - 3\hat{j}}{\sqrt{13}}$ and $\hat{i}_2 = \frac{C\hat{i} - 4\hat{j}}{\sqrt{C^2 + 16}}$
Interface is horizontal, so normal is $\hat{n} = \hat{j}$
Calculate $\sin\theta_1 = \frac{2}{\sqrt{13}}$ (x-component of incident ray)
Calculate $\sin\theta_2 = \frac{C}{\sqrt{C^2 + 16}}$ (x-component of refracted ray)
Some distant star is to be observed by some telescope of diameter of objective lens $a$, at an angular resolution of $3.0\times 10^{-7}$ radian. If the wavelength of light from the star reaching the telescope is 500 nm, the minimum diameter of the objective lens of the telescope is __________ cm. (nearest integer)
Show answer and solution
Answer: 203
Key idea: Rayleigh Criterion for Angular Resolution of Telescope
Write the Rayleigh criterion formula for angular resolution: $\theta = \frac{1.22\lambda}{a}$
