Ray Optics JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Ray Optics, free to read — no sign-in needed. The full chapter has 71 questions; sign in to attempt the remaining 66 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   A beam of polychromatic light passes through a thin prism of prism angle $6^\circ$. The refractive index of the material of the prism varies with wavelength $(\lambda)$ as $n(\lambda) = \alpha\lambda + \dfrac{\beta}{\lambda^2}$, where $\alpha = 3\ \mu\text{m}^{-1}$ and $\beta = 0.096\ \mu\text{m}^2$. If $\lambda_{\min}$ is the wavelength at which the angle of minimum deviation $D_m$ is smallest, then the correct value of $D_m$ at $\lambda_{\min}$ is
   1. A.$6.4^\circ$
   2. B.$4.8^\circ$
   3. C.$3.2^\circ$
   4. D.$2.4^\circ$
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   Answer: (B)

   For a thin prism, the angle of deviation is given by $D = (n - 1)A$. To find the smallest value of deviation $D_m$, we need to find the minimum value of the refractive index $n(\lambda)$. Given $n(\lambda) = \alpha\lambda + \dfrac{\beta}{\lambda^2}$. Differentiating $n(\lambda)$ with respect to $\lambda$ and equating to zero for minimum: $\dfrac{dn}{d\lambda} = \alpha - \dfrac{2\beta}{\lambda^3} = 0$ $\lambda^3 = \dfrac{2\beta}{\alpha}$ Substituting the given values $\alpha = 3\ \mu\text{m}^{-1}$ and $\beta = 0.096\ \mu\text{m}^2$: $\lambda^3 = \dfrac{2 \times 0.096}{3} = 0.064$ $\lambda = 0.4\ \mu\text{m}$ Now, substituting $\lambda = 0.4\ \mu\text{m}$ back into the expression for $n(\lambda)$: $n_{\min} = 3(0.4) + \dfrac{0.096}{(0.4)^2}$ $n_{\min} = 1.2 + \dfrac{0.096}{0.16} = 1.2 + 0.6 = 1.8$ The smallest angle of deviation is: $D_m = (n_{\min} - 1)A$ $D_m = (1.8 - 1) \times 6^\circ = 0.8 \times 6^\circ = 4.8^\circ$ Answer: $4.8^\circ$
2. Q2JEE Advanced Adv 2013 (Paper 1)
   The image of an object, formed by a plano - convex lens at a distance of 8 m behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is $\dfrac{2}{3}$ times the wavelength in free space. The radius of the curved surface of the lens is
   1. A.1 m
   2. B.2 m
   3. C.3 m
   4. D.6 m
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   Answer: (C)

   V = 8m Magnification =$\dfrac{\text{v}}{\text{u}}=\dfrac{-1}{3}$ u = -24 also $\mu =\dfrac{\text{c}}{{V}_{m}}=\dfrac{{\lambda }_{a}}{{\lambda }_{m}}=\dfrac{3}{2}$ $\dfrac{1}{\text{f}}=\left(\mu -1\right)\left(\dfrac{1}{{\text{R}}_{1}}-\dfrac{1}{{\text{R}}_{2 ​}}\right)$ $\dfrac{1}{\text{f}}=\left(\mu -1\right)\left(\dfrac{1}{\infty }-\dfrac{1}{-\text{R}}\right)$ $\dfrac{1}{\text{f}}=\left(\dfrac{3}{2}-1\right)\dfrac{1}{\text{R}}$ f = 2R from lens formula $\dfrac{1}{\text{v}}-\dfrac{1}{\text{u}}=\dfrac{1}{\text{f}}$ $\dfrac{1}{8}+\dfrac{1}{24}=\dfrac{1}{\text{2R}}$ $\dfrac{4}{\text{24}}=\dfrac{1}{\text{2R}}$ $\Rightarrow \text{R}=3\text{m}$
3. Q3JEE Advanced Adv 2013 (Paper 1)
   A ray of light, travelling in the direction $\dfrac{1}{2}\left(\hat{i}+\sqrt{3}\hat{j}\right)$, is incident on a plane mirror. After reflection, it travels along the direction $\dfrac{1}{2}\left(\hat{i}-\sqrt{3}\hat{j}\right)$. The angle of incidence is
   1. A.$30^{\circ}$
   2. B.$45^{\circ}$
   3. C.$60^{\circ}$
   4. D.$75^{\circ}$
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   Answer: (A)

   $\alpha$ = angle between vectors $\text{cos}\alpha =\dfrac{\dfrac{1}{2}\left(\hat{\text{i}}+\sqrt{3}\hat{\text{j}}\right)\cdot \dfrac{1}{2}\left(\hat{\text{i}}-\sqrt{3}\hat{\text{j}}\right)}{\left|\hat{\text{i}}+\sqrt{3}\hat{\text{j}}\right|\left|\hat{\text{i}}-\sqrt{3}\hat{\text{j}}\right|}$ $=\dfrac{1}{4}\dfrac{\left(1-3\right)}{1\times 1}=\dfrac{-2}{4}=-\dfrac{1}{2}$ $\alpha =120^{\circ}=\dfrac{2\pi }{3}$ $∴\pi -\alpha =2\theta$ $\pi =\dfrac{2\pi }{3}=2\theta$ $\theta =\dfrac{\pi }{3\times 2}=\dfrac{\pi }{6}=30^{\circ}$
4. Q4JEE Advanced Adv 2010 (Paper 2)
   Image of an object approaching a convex mirror of radius of curvature $20 \mathrm{~m}$ along its optical axis is observed to move from $\frac{25}{3} \mathrm{~m}$ to $\frac{50}{7} \mathrm{~m}$ in $30 \mathrm{~s}$. What is the speed of the object in $\mathrm{km} \mathrm{h}^{-1}$ ?
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   Answer: 3

   Using mirror formula twice, $\frac{1}{+25 / 3}+\frac{1}{-u_1}=\frac{1}{+10}$ or $\quad \frac{1}{u_1}=\frac{3}{25}-\frac{1}{10}$ or $u_1=50 \mathrm{~m}$ and, $\quad \frac{1}{(+50 / 7)}+\frac{1}{-u_2}=\frac{1}{+10}$ $\therefore \quad \frac{1}{u_2}=\frac{7}{50}-\frac{1}{10}$ or $u_2=25 \mathrm{~m}$ Speed of object $=\frac{u_1-u_2}{\text { time }}$ $=\frac{25}{30} \mathrm{~ms}^{-1}$ $=3 \mathrm{kmh}^{-1}$ $\therefore$ The answer is 3 .
5. Q5JEE Advanced Adv 2010 (Paper 1)
   The focal length of a thin biconvex lens is $20 \mathrm{~cm}$. When an object is moved from a distance of $25 \mathrm{~cm}$ in front of it to $50 \mathrm{~cm}$, the magnification of its image changes from $m_{25}$ to $m_{50}$. The ratio $\frac{m_{25}}{m_{50}}$ is
   Show answer & solution

   Answer: 6

   $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ or $$\frac{u}{v}-1=\frac{u}{f}$$ $\begin{array}{rlrl} & \text { or } & \frac{u}{v} & =\left(\frac{u+f}{f}\right) \\ \therefore & m=\frac{v}{u} & =\left(\frac{f}{u+f}\right) \\ \frac{m_{25}}{m_{50}} & =\frac{\left(\frac{20}{-25+20}\right)}{\left(\frac{20}{-50+20}\right)}=6\end{array}$ $\therefore$ answer is 6 .