Ray & Wave Optics — Previous Year Questions (JEE Main 2026)

For single-slit diffraction, the $n^{\text{th}}$ minimum occurs at

$$a\sin\theta = n\lambda \implies y_n = \frac{n\lambda D}{a}\quad(\text{small angles}).$$

Positions of the $1^{\text{st}}$ and $3^{\text{rd}}$ minima:

$$y_1 = \frac{\lambda D}{a},\qquad y_3 = \frac{3\lambda D}{a}.$$

Linear separation:

$$\Delta y = y_3 - y_1 = \frac{3\lambda D}{a} - \frac{\lambda D}{a} = \frac{2D\lambda}{a}.$$

Answer: C ($2\frac{D\lambda}{a}$)

The angular resolution (limit) of a telescope is

$$\theta = \frac{1.22\,\lambda}{R} \implies R = \frac{1.22\,\lambda}{\theta}.$$

Substitute $\lambda = 500\times10^{-9}\ \text{m}$ and $\theta = 5\times10^{-7}\ \text{rad}$:

$$R = \frac{1.22 \times 500\times10^{-9}}{5\times10^{-7}} = \frac{6.1\times10^{-7}}{5\times10^{-7}} = 1.22\ \text{m}.$$$$R = 1.22\ \text{m} = 122\ \text{cm}.$$

Answer: B (122)

Phase difference from path difference: $\phi = \dfrac{2\pi}{\lambda}\,\Delta x$, and for two equal-amplitude waves

$$I = 4I_0\cos^2\!\left(\frac{\phi}{2}\right).$$

At $A$: $\Delta x = \dfrac{\lambda}{3} \Rightarrow \phi_A = \dfrac{2\pi}{3} \Rightarrow \dfrac{\phi_A}{2} = 60^\circ$

$$I_A = 4I_0\cos^2 60^\circ = 4I_0 \cdot \tfrac{1}{4} = I_0.$$

At $B$: $\Delta x = \dfrac{\lambda}{6} \Rightarrow \phi_B = \dfrac{\pi}{3} \Rightarrow \dfrac{\phi_B}{2} = 30^\circ$

$$I_B = 4I_0\cos^2 30^\circ = 4I_0 \cdot \tfrac{3}{4} = 3I_0.$$

Ratio:

$$\frac{I_A}{I_B} = \frac{I_0}{3I_0} = \frac{1}{3}.$$

Answer: C ($1/3$)

Unpolarized light through the polarizer:

$$I_1 = \frac{I_0}{2}.$$

The optically active solution rotates the plane of polarization by angle $\theta$. Since the analyser is parallel to the polarizer ($0^\circ$), the angle between the (rotated) polarization plane and the analyser axis is $\theta$. By Malus’s law:

$$I = I_1\cos^2\theta = \frac{I_0}{2}\cos^2\theta.$$

Set equal to the given output:

$$\frac{I_0}{2}\cos^2\theta = \frac{3}{8}I_0 \implies \cos^2\theta = \frac{3}{4} \implies \cos\theta = \frac{\sqrt3}{2}.$$$$\theta = 30^\circ.$$

Answer: 30

For a thin prism, the minimum deviation is

$$\delta_m = (\mu - 1)A = (1.5 - 1)A = 0.5\,A.$$

At minimum deviation the ray passes symmetrically, so the incident angle is

$$i = \frac{A + \delta_m}{2} = \frac{A + 0.5A}{2} = 0.75\,A.$$

Therefore the required ratio:

$$\frac{i}{\delta_m} = \frac{0.75A}{0.5A} = \frac{3}{2}.$$

Answer: B ($3 : 2$)

The full angular width of the central maximum is

$$\Delta\theta = \frac{2\lambda}{a}.$$

Substitute $\lambda = 628\times10^{-9}\ \text{m}$ and $a = 0.2\times10^{-3}\ \text{m}$:

$$\Delta\theta = \frac{2 \times 628\times10^{-9}}{0.2\times10^{-3}} = 6.28\times10^{-3}\ \text{rad}.$$

Convert to degrees:

$$\Delta\theta = 6.28\times10^{-3} \times \frac{180}{\pi} \approx 0.36^\circ = 36 \times 10^{-2}\ \text{degrees}.$$$$\alpha = 36.$$

Answer: 36

Equiconvex lens: $R_1 = +R$, $R_2 = -R$. Lens maker’s formula:

$$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (\mu - 1)\left(\frac{1}{R} + \frac{1}{R}\right) = \frac{2(\mu - 1)}{R}.$$

Hence

$$\frac{f}{R} = \frac{1}{2(\mu - 1)} = \frac{1}{2(1.4 - 1)} = \frac{1}{0.8} = 1.25.$$

Answer: D (1.25)

Let the intensity after the first polarizer (at $30^\circ$) be $I_1 = \dfrac{I}{2}$.

Without the third polarizer (axes $30^\circ \to 90^\circ$, i.e. $60^\circ$ apart):

$$I_{\text{out}} = I_1\cos^2 60^\circ = \frac{I}{2}\cdot\frac{1}{4} = \frac{I}{8}.$$

With the third polarizer at $60^\circ$ ($30^\circ \to 60^\circ \to 90^\circ$, each step $30^\circ$):

$$I_{\text{out}}' = I_1\cos^2 30^\circ \cos^2 30^\circ = \frac{I}{2}\cdot\frac{3}{4}\cdot\frac{3}{4} = \frac{9I}{32}.$$

Ratio (with $:$ without):

$$\frac{I_{\text{out}}'}{I_{\text{out}}} = \frac{9I/32}{I/8} = \frac{9}{32}\times 8 = \frac{9}{4}.$$

Answer: C (9/4)

Introducing a sheet of refractive index $\mu$ and thickness $t$ adds an extra path $(\mu - 1)t$. The central fringe shifts by $n$ fringes when

$$(\mu - 1)t = n\lambda.$$

Here $n = 7$, $\mu = 1.56$, $\lambda = 450\times10^{-9}\ \text{m}$:

$$t = \frac{n\lambda}{\mu - 1} = \frac{7 \times 450\times10^{-9}}{1.56 - 1} = \frac{3150\times10^{-9}}{0.56}.$$$$t = 5625\times10^{-9}\ \text{m}.$$$$\alpha = 5625.$$

Answer: 5625

For two equal sources, $I = I_{\max}\cos^2\!\left(\dfrac{\phi}{2}\right)$. Given $I = \dfrac{3}{4}I_{\max}$:

$$\cos^2\!\left(\frac{\phi}{2}\right) = \frac{3}{4} \implies \cos\frac{\phi}{2} = \frac{\sqrt3}{2} \implies \frac{\phi}{2} = 30^\circ \implies \phi = 60^\circ = \frac{\pi}{3}.$$

Path difference:

$$\Delta x = \frac{\lambda}{2\pi}\,\phi = \frac{\lambda}{2\pi}\cdot\frac{\pi}{3} = \frac{\lambda}{6}.$$

Comparing with $\Delta x = \dfrac{\lambda}{x}$ gives $x = 6$.

Answer: 6

Sunlight is effectively parallel, so it converges at the focal point. The shortest burning time occurs when the paper is at the focus:

$$f = 30\ \text{cm}.$$

For an equiconvex lens with both surfaces of radius $R = 60\ \text{cm}$:

$$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R} + \frac{1}{R}\right) = \frac{2(\mu - 1)}{R}.$$

Substitute $f = 30$, $R = 60$:

$$\frac{1}{30} = \frac{2(\mu - 1)}{60} = \frac{\mu - 1}{30} \implies \mu - 1 = 1 \implies \mu = 2.$$

So $\dfrac{\alpha}{10} = 2 \implies \alpha = 20$.

Answer: 20

For an equilateral prism, $A = 60^\circ$ and given $\delta_m = \dfrac{A}{2} = 30^\circ$.

Prism formula:

$$\mu = \frac{\sin\!\left(\dfrac{A + \delta_m}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)} = \frac{\sin\!\left(\dfrac{60^\circ + 30^\circ}{2}\right)}{\sin 30^\circ} = \frac{\sin 45^\circ}{\sin 30^\circ}.$$$$\mu = \frac{1/\sqrt2}{1/2} = \frac{2}{\sqrt2} = \sqrt2 \approx 1.414.$$

Answer: C ($\sqrt{2}$)

A concave mirror can give a magnified image of magnitude $2$ in two ways: a real inverted image ($m = -2$) and a virtual erect image ($m = +2$).

Using $m = \dfrac{f}{f - u}$ (with $f = -10\ \text{cm}$, distances measured with sign):

Real image $m = -2$: $\;-2 = \dfrac{-10}{-10 - u} \Rightarrow -10 - u = 5 \Rightarrow u = -15\ \text{cm}$ (object at 15 cm).

Virtual image $m = +2$: $\;+2 = \dfrac{-10}{-10 - u} \Rightarrow -10 - u = -5 \Rightarrow u = -5\ \text{cm}$ (object at 5 cm).

Distance between the two object positions:

$$|{-15}| - |{-5}| = 15 - 5 = 10\ \text{cm}.$$

Answer: 10

Fringe width $\beta = \dfrac{\lambda D}{d}$. In a medium the wavelength reduces to $\lambda' = \dfrac{\lambda}{\mu}$, so

$$\beta' = \frac{\beta}{\mu} = \frac{2.4\ \mu\text{m}}{1.2} = 2\ \mu\text{m}.$$

Answer: B ($2$)

Refractive index of the prism material:

$$\mu = \frac{c}{v} = \frac{3\times10^8}{2.12\times10^8} \approx 1.414 = \sqrt2.$$

For an equilateral prism $A = 60^\circ$:

$$\mu = \frac{\sin\!\left(\dfrac{A + \delta_m}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)} \implies \sin\!\left(\frac{60^\circ + \delta_m}{2}\right) = \mu\sin 30^\circ = \sqrt2 \cdot \frac{1}{2} = \frac{1}{\sqrt2}.$$$$\frac{60^\circ + \delta_m}{2} = 45^\circ \implies 60^\circ + \delta_m = 90^\circ \implies \delta_m = 30^\circ.$$

Answer: B ($30$)

For a symmetric biconvex lens (both radii $R$):

$$\frac{1}{f_{\text{bi}}} = (\mu - 1)\frac{2}{R}.$$

Cutting it vertically gives a plano-convex lens (one flat face, $R \to \infty$):

$$\frac{1}{f_{\text{pc}}} = (\mu - 1)\frac{1}{R} = \frac{1}{2f_{\text{bi}}} \implies f_{\text{pc}} = 2f_{\text{bi}}.$$

The objective’s linear magnification is $m_o \approx \dfrac{L}{f_o}$, where $L$ is the tube length. To keep $m_o$ (hence the overall magnification) unchanged with the object distance fixed, when $f_o$ doubles the tube length must also double:

$$\frac{L'}{f_{\text{pc}}} = \frac{L}{f_{\text{bi}}} \implies L' = \frac{f_{\text{pc}}}{f_{\text{bi}}}L = 2L.$$

Answer: A (increased two times)

For lenses in contact, powers add:

$$\frac{1}{f} = \frac{1}{f_{\text{convex}}} - \frac{1}{|f_{\text{concave}}|}.$$

The combination behaves as a concave lens when $\dfrac{1}{f} < 0$, i.e.

$$\frac{1}{f_{\text{convex}}} < \frac{1}{|f_{\text{concave}}|} \implies |f_{\text{convex}}| > |f_{\text{concave}}|.$$

(Interchanging the order of two thin lenses in contact does not change the combined focal length, so that option is incorrect.)

Answer: A (behaves as concave lens if $|f_{convex}| > |f_{concave}|$)

Lens $P$ ($f = +10$, $u = -15$):

$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{15} = \frac{1}{30} \implies v = +30\ \text{cm}.$$

Image forms 30 cm to the right of $P$; $m_P = \dfrac{v}{u} = \dfrac{30}{-15} = -2$.

Lens $Q$ is 15 cm right of $P$, so this image lies $30 - 15 = 15\ \text{cm}$ to the right of $Q$ — a virtual object, $u = +15$, $f = +15$:

$$\frac{1}{v} = \frac{1}{15} + \frac{1}{15} = \frac{2}{15} \implies v = +7.5\ \text{cm}.$$

Real image, 7.5 cm right of $Q$; $m_Q = \dfrac{v}{u} = \dfrac{7.5}{15} = +0.5$.

Total magnification:

$$m = m_P \cdot m_Q = (-2)(0.5) = -1.$$

Image is real, 7.5 cm right of $Q$, same size as $AB$ (inverted).

Answer: B (real, formed at 7.5 cm right of lens $Q$, with a size same as that of $AB$)

A point exactly opposite one slit is at $y = \dfrac{d}{2}$ from the centre. Path difference:

$$\Delta x = \frac{d\,y}{D} = \frac{d\cdot (d/2)}{10d} = \frac{d}{20} = \frac{5\lambda}{20} = \frac{\lambda}{4}.$$

Phase difference:

$$\phi = \frac{2\pi}{\lambda}\cdot\frac{\lambda}{4} = \frac{\pi}{2}.$$

Intensity:

$$I = I_0\cos^2\!\left(\frac{\phi}{2}\right) = I_0\cos^2\!\left(\frac{\pi}{4}\right) = I_0\cdot\frac{1}{2} = \frac{I_0}{2}.$$

Answer: B ($\dfrac{I_0}{2}$)

A silvered lens behaves as an equivalent mirror. Light passes through the lens, reflects off the silvered surface, and passes through the lens again, so

$$P_{\text{eq}} = 2P_{\text{lens}} + P_{\text{mirror}}.$$

Lens power (biconvex, $R = 20$):

$$P_{\text{lens}} = (\mu-1)\left(\frac{1}{R}+\frac{1}{R}\right) = 0.5\cdot\frac{2}{20} = \frac{1}{20}\ \text{cm}^{-1}.$$

Mirror power (silvered surface, $R = 20$, so $f_m = R/2 = 10$):

$$P_{\text{mirror}} = \frac{1}{f_m} = \frac{2}{R} = \frac{1}{10}\ \text{cm}^{-1}.$$

Equivalent power and focal length:

$$P_{\text{eq}} = 2\cdot\frac{1}{20} + \frac{1}{10} = \frac{1}{10}+\frac{1}{10} = \frac{1}{5} \implies f_{\text{eq}} = 5\ \text{cm}.$$

For a mirror, object and image coincide when the object is at the centre of curvature, i.e. at $2f_{\text{eq}}$:

$$u = 2f_{\text{eq}} = 2\times 5 = 10\ \text{cm}.$$

Answer: A (10)

The angular resolution limit of a telescope is

$$\theta = \frac{1.22\,\lambda}{a} \implies a = \frac{1.22\,\lambda}{\theta}.$$

Substitute $\lambda = 500\times10^{-9}\ \text{m}$ and $\theta = 3.0\times10^{-7}\ \text{rad}$:

$$a = \frac{1.22 \times 500\times10^{-9}}{3.0\times10^{-7}} = \frac{6.1\times10^{-7}}{3.0\times10^{-7}} \approx 2.033\ \text{m}.$$$$a \approx 203.3\ \text{cm} \approx 203\ \text{cm}.$$

Answer: 203