JEE Advanced 2023 — Paper 2 — PHY-II solution

JEE Advanced2023Paper 2PHY-II

A monochromatic light wave is incident normally on a glass slab of thickness $d$ , as shown in the figure. The refractive index of the slab increases linearly from $n_{1}$ to $n_{2}$ over the height $h$ . Which of the following statement(s) is (are) true about the light wave emerging out of the slab?

A
It will deflect up by an angle $tan^{- 1} [\frac{( n _{2}^{2} - n _{1}^{2} ) d}{2 h}]$
B
It will deflect up by an angle $tan^{- 1} [\frac{( n _{2} - n _{1} ) d}{h}]$
C
It will not deflect.
D
The deflection angle depends only on $(n_{2} - n_{1})$ and not on the individual values of $n_{1}$ and $n_{2}$ .

Solution

Consider two extreme rays of the same incident wavefront: one travelling along the top edge of the slab (refractive index $n_{2}$ ) and one along the bottom edge ( $n_{1}$ ). Both traverse the same geometric thickness $d$ , but they accumulate different optical path lengths $n_{2} d$ and $n_{1} d$ .

The top ray is retarded relative to the bottom ray. After exit, the wavefront tilts: along the height $h$ of the wavefront, there is a sideways shift $ℓ$ given by matching optical paths,

$n_{1} d + ℓ = n_{2} d \Rightarrow ℓ = (n_{2} - n_{1}) d$ .

The tilt angle $θ$ of the emergent wavefront with respect to the original direction satisfies

$tan θ = \frac{ℓ}{h} = \frac{( n _{2} - n _{1} ) d}{h}$ .

Since the top of the slab has the higher index, the wave bends upward, towards the denser region. The result depends solely on the difference $(n_{2} - n_{1})$ , justifying option (D) as well.

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