Ray Optics JEE Main previous year questions with solutions

5 solved JEE Main questions on Ray Optics, free to read — no sign-in needed. The full chapter has 288 questions; sign in to attempt the remaining 283 in the exam simulator.

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1. Q1JEE Main 2026 (23 Jan, Shift 1)Critical angle and total internal reflection
   Consider light travelling from a medium $A$ to medium $B$ separated by a plane interface. If the light undergoes total internal reflection during its travel from medium $A$ to $B$ and the speed of light in media $A$ and $B$ are $2.4 \times 10^{8} \mathrm{~m} / \mathrm{s}$ and $2.7 \times 10^{8} \mathrm{~m} / \mathrm{s}$, respectively, then the value of critical angle is :
   1. A.$\cos ^{-1}\left(\frac{8}{9}\right)$
   2. B.$\sin ^{-1}\left(\frac{9}{8}\right)$
   3. C.$\cot ^{-1}\left(\frac{3}{\sqrt{13}}\right)$
   4. D.$\tan ^{-1}\left(\frac{8}{\sqrt{17}}\right)$
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   Answer: (D)

   The refractive index of a medium is inversely proportional to the speed of light in that medium, given by $n = \frac{c}{v}$. For light travelling from medium $A$ to medium $B$, the critical angle $\theta_c$ is defined when light travels from a denser to a rarer medium. Given speeds are $v_A = 2.4 \times 10^8$ m/s and $v_B = 2.7 \times 10^8$ m/s. The refractive indices are $n_A = \frac{c}{v_A}$ and $n_B = \frac{c}{v_B}$. The critical angle $\theta_c$ is given by $\sin \theta_c = \frac{n_B}{n_A} = \frac{v_A}{v_B}$. Substituting the values: $\sin \theta_c = \frac{2.4 \times 10^8}{2.7 \times 10^8} = \frac{24}{27} = \frac{8}{9}$. To find the value in terms of $\tan^{-1}$, we use the trigonometric identity for a right-angled triangle where perpendicular $p = 8$ and hypotenuse $h = 9$. The base $b = \sqrt{h^2 - p^2} = \sqrt{9^2 - 8^2} = \sqrt{81 - 64} = \sqrt{17}$. Thus, $\tan \theta_c = \frac{p}{b} = \frac{8}{\sqrt{17}}$. Therefore, $\theta_c = \tan^{-1}\left(\frac{8}{\sqrt{17}}\right)$.
2. Q2JEE Main 2025 (07 Apr, Shift 1)Lens Formula, Refraction of light from curved surface
   Two thin convex lenses of focal length 30 cm and 10 cm are placed coaxially, 10 cm apart. The power of this combination is :
   1. A.5 D
   2. B.1 D
   3. C.20 D
   4. D.10 D
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   Answer: (D)

   $\mathrm{f}_1=30 \mathrm{~cm}, \mathrm{f}_2=10 \mathrm{~cm}$ $\frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2}-\frac{\mathrm{d}}{\mathrm{f}_1 \mathrm{f}_2}, \mathrm{~d}=$ distance between lens $\frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{1}{0.3}+\frac{1}{0.1}-\frac{0.1}{(0.3)(0.1)}$ $\frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{1}{0.1}$ Power $=\frac{1}{\mathrm{f}_{\mathrm{eq}}}=10 \mathrm{D}$
3. Q3JEE Main 2024 (29 Jan, Shift 1)Mirror Formula, Reflection in Spherical Mirror
   A convex mirror of radius of curvature $30cm$ forms an image that is half the size of the object. The object distance is :
   1. A.$-45cm$
   2. B.$45cm$
   3. C.$-15cm$
   4. D.$15cm$
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   Answer: (C)

   Given $R=30cm$ So, $f=\dfrac{R}{2}=+15cm$ For convex mirror, virtual image is formed for real object. Thus, magnification $\left(m\right)=\dfrac{1}{2}$ The magnification can be written as $m=-\dfrac{v}{u}...\left(1\right)$ Equation (1) implies that $\dfrac{1}{2}=-\dfrac{v}{u} \Rightarrow v=-\dfrac{u}{2}$ The relation between the object distance, image distance and the focal length of the mirror is given by $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}...\left(2\right)$ Equation (2) implies that $\dfrac{1}{15}=\dfrac{1}{-\dfrac{u}{2}}+\dfrac{1}{u} =-\dfrac{1}{u} \Rightarrow u=-15cm$
4. Q4JEE Main 2023 (12 Apr, Shift 1)Combination of Lens and Mirrors
   Two convex lenses of focal length $20cm$ each are placed coaxially with a separation of $60cm$ between them. The image of the distant object formed by the combination is at _____ $cm$ from the first lens.
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   Answer: 100

   Let ${L}_{1}$ be the first lens and ${L}_{2}$ be the second lens. The first refraction is in ${L}_{1}$. Using lens formula, ${u}_{1}=\infty ,{f}_{1}=20cm, \dfrac{1}{{v}_{1}}-0=\dfrac{1}{{f}_{1}} \Rightarrow {v}_{1}=20cm$ Hence, the distance between the second image and the ${L}_{2}$ would be $\left(60-20\right)=40cm$. For the second lens ${L}_{2}$, ${u}_{2}=-40cm,{f}_{2}=20cm, \dfrac{1}{{v}_{2}}-\dfrac{1}{{u}_{2}}=\dfrac{1}{{f}_{2}} \Rightarrow \dfrac{1}{{v}_{2}}+\dfrac{1}{40}=\dfrac{1}{20} \Rightarrow \dfrac{1}{{v}_{2}}=\dfrac{1}{40} \Rightarrow {v}_{2}=40cm$ So, the distance of image from first lens is $x=40cm+60cm$ $=100cm$
5. Q5JEE Main 2022 (28 Jul, Shift 1)Optical Instruments
   In normal adjustment, for a refracting telescope, the distance between objective and eye piece is $30cm$. The focal length of the objective, when the angular magnification of the telescope is $2$, will be:
   1. A.$20cm$
   2. B.$30cm$
   3. C.$10cm$
   4. D.$15cm$
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   Answer: (A)

   Let ${f}_{0}$ be the focal length of objective and ${f}_{e}$ is the focal length of eyepiece. Given here, ${f}_{0}+{f}_{e}=30$. Angular magnification of telescope is given by $m=\dfrac{{f}_{0}}{{f}_{e}}$. So, $2=\dfrac{{f}_{0}}{{f}_{e}}\Rightarrow {f}_{0}=2{f}_{e}$ Then, we have ${f}_{0}+\dfrac{{f}_{0}}{2}=30\Rightarrow \dfrac{3{f}_{0}}{2}=30$ Or, ${f}_{0}=20cm$.