NEET Wave Optics Previous Year Questions with Solutions

Brewster’s law states that when light is incident at a polarizing angle i_B, the tangent of the polarizing angle is equal to the refractive index μ of the reflecting medium:

• • By the fundamental law of reflection, angle of reflection equals angle of incidence: r' = i_B = 60°.

At Brewster’s angle, the reflected and refracted rays are mutually perpendicular (i_B + r = 90°). Electrons oscillating parallel to the refracted ray cannot radiate along their axis of oscillation towards the reflected direction.

• • Therefore, the reflected light contains only vibrations perpendicular to the plane of incidence (s-polarization) and is completely plane polarized.
• • The transmitted light contains both parallel and residual perpendicular components, making it only partially polarized.

The light emerging from the first polaroid P₁ has intensity I₀ and is linearly polarized along P₁’s transmission axis. When it strikes polaroid P₂ at an angle θ = 22.5°:

• • The light transmitted through P₂ is now polarized along P₂’s axis.

P₃ is crossed with P₁, meaning its transmission axis is at 90° to P₁. The angle between P₂ and P₃ is therefore 90° - 22.5° = 67.5°:

• • Since cos(67.5°) = cos(90° - 22.5°) = sin(22.5°):
• • I_{final} = I_0 [\cos(22.5^\circ) \sin(22.5^\circ)]^2.

Using the identity 2 sin θ cos θ = sin(2θ):

• • Matches Option (3) exactly.

In Young’s double slit experiment, the path difference between light waves emerging from slits S1 and S2 arriving at the central point O on the perpendicular bisector is identically zero:

Since Δx = 0 satisfies the condition for constructive interference (Δx = nλ with n = 0) for every wavelength in the visible spectrum simultaneously, all color components (VIBGYOR) constructively combine with maximum intensity. Their additive combination yields a bright white central fringe.

For points away from the center (Δx ≠ 0), the fringe width β depends directly on wavelength:

Because λ_violet ≈ 400 nm is smaller than λ_red ≈ 700 nm, the violet bright fringes form closest to the central white fringe, with red forming farther out. After a few colored fringes, the overlapping of different bright and dark bands of different colors produces uniform white illumination.

When an unpolarised light beam strikes a transparent refracting medium at Brewster’s angle i_B, the angle of incidence satisfies:

By Snell’s law, sin i_B / sin r = μ = sin i_B / cos i_B, which implies cos i_B = sin r = cos(90° - r). Therefore:

The oscillating electric dipoles in the medium produce the reflected wave. Because dipoles cannot radiate electromagnetic energy along their oscillation axes, the electric field oscillations parallel to the plane of incidence are completely absent in the reflected ray. Thus, the reflected light is completely (100%) linearly polarised with electric vectors perpendicular to the plane of incidence (dots).

The refracted ray transmits into the medium. Since only a fraction of the perpendicular component is reflected, the refracted beam contains the entire parallel component and the remaining fraction of the perpendicular component. It is therefore partially polarised.

In Fraunhofer single-slit diffraction produced by a slit of width a, the condition for the first diffraction minimum is given by a sin θ = λ. For small diffraction angles (sin θ ≈ θ in radians), the angular width of the central maximum is 2θ = 2λ/a, and the angular separation of adjacent fringes is:

Linear fringe separation on the screen is y = D θ = λ D / a, which is directly proportional to D. However, the ANGULAR separation θ depends strictly on the ratio λ/a and is completely independent of the distance D between the slit plane and the screen:

From the governing formula θ = λ / a, the angular separation is directly proportional to the wavelength of light used:

Statement I correctly asserts that angular separation does not change when the screen is moved away. Statement II incorrectly claims that angular separation decreases with increasing wavelength (it actually increases). Therefore, Statement I is true and Statement II is false.

When unpolarized light is incident on the interface separating two media, the reflected light is completely plane-polarized when the angle of incidence equals Brewster's angle ($$i_b$$), where reflected and refracted rays are mutually perpendicular ($$i_b + r = 90^\circ$$).

Using Snell's law with $$r = 90^\circ - i_b$$: $$\frac{\sin i_b}{\sin r} = \mu \implies \frac{\sin i_b}{\sin(90^\circ - i_b)} = \frac{\sin i_b}{\cos i_b} = \tan i_b = \mu$$

For any transparent optical medium (e.g. water $$\mu = 1.33$$, glass $$\mu = 1.5$$, diamond $$\mu = 2.42$$) surrounded by air/vacuum ($$\mu_{\text{air}} = 1$$): $$\mu > 1$$ And for any real physical material, refractive index is finite: $$1 < \mu < \infty$$

Because the tangent function is strictly monotonically increasing in the first quadrant ($$0^\circ < i_b < 90^\circ$$): $$\tan(45^\circ) = 1 \quad \text{and} \quad \tan(90^\circ) = \infty$$ $$1 < \tan i_b < \infty \implies 45^\circ < i_b < 90^\circ$$ This confirms Option (3).

In Young's double-slit experiment, the fringe width $$\beta$$ (distance between consecutive bright or dark fringes) is given by: $$\beta = \frac{\lambda D}{d}$$, where $$\lambda$$ is the wavelength of light, $$D$$ is the distance between slits and screen, and $$d$$ is the slit separation.

- Separation between coherent sources is halved: $$d' = \frac{d}{2}$$ - Distance of the screen is doubled: $$D' = 2D$$ - Wavelength $$\lambda$$ remains unchanged.

$$\beta' = \frac{\lambda D'}{d'} = \frac{\lambda (2D)}{\frac{d}{2}} = \frac{2}{\frac{1}{2}} \left(\frac{\lambda D}{d}\right) = 4 \left(\frac{\lambda D}{d}\right) = 4\beta$$.

Thus, the fringe width becomes four times its original value, matching Option 3.

Light passing through a circular telescope objective of diameter $$D$$ forms an Airy diffraction pattern consisting of a central bright disc surrounded by dark and bright concentric rings. The angular radius of the first dark ring (Airy disc) defines the angular limit of resolution: $$\Delta\theta = \frac{1.22 \lambda}{D}$$.

- Wavelength of light: $$\lambda = 600\text{ nm} = 600 \times 10^{-9}\text{ m} = 6 \times 10^{-7}\text{ m}$$ - Diameter of telescope objective: $$D = 2\text{ m}$$.

$$\Delta\theta = \frac{1.22 \times (6 \times 10^{-7}\text{ m})}{2\text{ m}} = 1.22 \times 3 \times 10^{-7} = 3.66 \times 10^{-7}\text{ rad}$$.

Matches Option 1 exactly.