Wave Optics — NEET UG practice

40 questions

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Sample questions with solutions

Q1 · 2026

In Young's double slit experiment, using monochromatic light of wavelength $\lambda$ , the intensity of light at a point on the screen where the path difference is $\lambda$ , is $K$ units. The intensity of light at a point where the path difference is $\frac{\lambda}{3}$ will be

A.
$\frac{K}{4}$
B.
$K$
C.
$2K$
D.
$\frac{K}{2}$

Answer: A

Recall that when two coherent waves of equal intensity $I_0$ superpose, the resultant intensity depends on the phase difference $\Delta\phi$ between them.

I = I_0\cos^2\left(\frac{\Delta\phi}{2}\right)

The phase difference $\Delta\phi$ is related to the path difference $\Delta x$ by the standard relation.

\Delta\phi = \frac{2\pi}{\lambda}\Delta x

Given that a path difference of $\lambda$ produces intensity $K$ , substitute $\Delta x = \lambda$ into the intensity formula.

K = I_0\cos^2\left(\frac{2\pi}{\lambda}\cdot\frac{\lambda}{2}\right) = I_0\cos^2(\pi) = I_0

Therefore $I_0 = K$ .

Now find the intensity for a path difference of $\frac{\lambda}{3}$ using the same formula.

I_1 = I_0\cos^2\left(\frac{2\pi}{\lambda}\cdot\frac{\lambda}{6}\right) = I_0\cos^2\left(\frac{\pi}{3}\right)

Since $\cos\left(\frac{\pi}{3}\right) = \frac12$ , simplify the expression.

I_1 = I_0\times\frac14 = \frac{K}{4}

Hence, the answer is A.

Q2 · 2026

In interference and diffraction, the light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe.

A. As there is no gain or loss of energy, these phenomena are consistent with the principle of conservation of energy.

B. Diffraction and interference are characteristics exhibited only by light waves.

Choose the correct answer from the options given below:

A.
$A$ is true and $B$ is also true
B.
$A$ is false, but $B$ is true
C.
$A$ is true, but $B$ is false
D.
Both A and B are false

Answer: C

Recall the principle of conservation of energy applied to wave superposition: the energy that disappears from a dark fringe region does not vanish, it simply reappears as extra energy at a bright fringe.

\text{Total energy before superposition} = \text{Total energy after superposition}

Given this matches exactly what Statement A describes, Statement A is true.

Recall that interference and diffraction are general properties of all wave motion, not something unique to light. Given that sound waves, water waves, and other mechanical waves also show interference and diffraction, Statement B is incorrect in restricting these phenomena to light alone.

Hence, the answer is C.

Q3 · 2025

An unpolarized light beam travelling in air is incident on a medium of refractive index 1.73 at Brewster's angle. Then

A.
Both reflected and transmitted light are perfectly polarized with angles of reflection and refraction close to $60^{\circ}$ and $30^{\circ}$ , respectively
B.
Transmitted light is completely polarized with angle of refraction close to $30^{\circ}$
C.
Reflected light is completely polarized and the angle of reflection is close to $60^{\circ}$
D.
Reflected light is partially polarized and the angle of reflection is close to $30^{\circ}$

Answer: C

Recall Brewster's law, which connects the refractive index $\mu$ of the medium to the polarising angle $\theta_p$ at which the reflected light becomes completely polarized.

\mu = \tan\theta_p

Substitute the given refractive index $\mu = 1.73 \approx \sqrt3$ into this relation.

\tan\theta_p = \sqrt3 \implies \theta_p = 60^{\circ}

Recall that at Brewster's angle, the reflected ray is completely (perfectly) polarized, while the transmitted (refracted) ray is only partially polarized. Given $\theta_p = 60^{\circ}$ is the angle of reflection here, the reflected light is completely polarized at an angle close to $60^{\circ}$ .

Hence, the answer is C.

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