JEE Challenger
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Deflection of Monochromatic Light Wave Through Gradient Slab

A monochromatic light wave is incident normally on a glass slab of thickness dd, as shown in the figure. The refractive index of the slab increases linearly from n1n_1 to n2n_2 over the height hh. Which of the following statement(s) is(are) true about the light wave emerging out of the slab?

Question Diagram 1

Options

A

It will deflect up by an angle tan1[(n22n12)d2h]\tan^{-1} \left[ \frac{(n_2^2 - n_1^2)d}{2h} \right].

B

It will deflect up by an angle tan1[(n2n1)dh]\tan^{-1} \left[ \frac{(n_2 - n_1)d}{h} \right].

Correct
C

It will not deflect.

D

The deflection angle depends only on (n2n1)(n_2 - n_1) and not on the individual values of n1n_1 and n2n_2.

Correct

Step-by-Step Solution

To determine the deflection of the monochromatic light wave emerging from the glass slab, we analyze the optical path length and wavefront orientation as the wave traverses the medium.

1. Refractive Index Profile

Let yy be the vertical coordinate measured from the bottom of the slab (y=0y = 0) to the top (y=hy = h). The refractive index n(y)n(y) increases linearly from n1n_1 to n2n_2: n(y)=n1+(n2n1h)yn(y) = n_1 + \left( \frac{n_2 - n_1}{h} \right) y

2. Optical Path Length (OPL)

A plane wave is incident normally on the left face of the slab. The optical path length traversed by the light wave at height yy through the slab of thickness dd is given by: OPL(y)=n(y)d=[n1+(n2n1h)y]d\text{OPL}(y) = n(y) \cdot d = \left[ n_1 + \left( \frac{n_2 - n_1}{h} \right) y \right] d

3. Wavefront Orientation and Deflection Angle

The phase ϕ(y)\phi(y) of the wave emerging at the right face of the slab is: ϕ(y)=ϕ02πλ0OPL(y)\phi(y) = \phi_0 - \frac{2\pi}{\lambda_0} \text{OPL}(y)

Since n2>n1n_2 > n_1, light at the top of the slab (y=hy = h) travels through a region of higher refractive index, meaning its phase velocity is slower compared to the light at the bottom (y=0y = 0). Consequently, the wave at the top experiences a larger phase delay, causing the emerging wavefront to tilt backward at the top.

The direction of wave propagation (the ray direction) is always perpendicular to the wavefront. The tangent of the deflection angle θ\theta relative to the horizontal (xx-axis) is given by the gradient of the optical path length along the yy-direction: tanθ=d(OPL)dy=ddndy=(n2n1)dh\tan \theta = \frac{d(\text{OPL})}{dy} = d \frac{dn}{dy} = \frac{(n_2 - n_1)d}{h}

Since dndy>0\frac{dn}{dy} > 0, the wave vector tilts upward. Therefore, the deflection angle is: θ=tan1[(n2n1)dh](deflected upwards)\theta = \tan^{-1} \left[ \frac{(n_2 - n_1)d}{h} \right] \quad \text{(deflected upwards)}

4. Conclusion and Option Analysis

  • Option A: Incorrect.
  • Option B: Correct. The light wave deflects up by an angle tan1[(n2n1)dh]\tan^{-1} \left[ \frac{(n_2 - n_1)d}{h} \right].
  • Option C: Incorrect.
  • Option D: Correct. The angle θ\theta depends only on the difference (n2n1)(n_2 - n_1) and the geometric parameters dd and hh, and not on the individual values of n1n_1 and n2n_2.

Correct Answer: (B) and (D)

Deflection of Monochromatic Light Wave Through Gradient Slab | Physics PYQ Solution - JEE Challenger