Name: Applications of TIR: Mirage and Looming
Uploaded: 2019-07-19
Duration: 4 min 7 s
Description: Let’s learn about mirages and looming in this interesting video. These are fascinating phenomena of light observed on hot summer days.

Question

Discuss refraction through a glass slab and show that emergent ray is parallel to the incident ray but displaced.

Accepted Answer

Let a light ray of light AB falls on the air-glass interface making an angle $(i)$ with the normal drawn at the surface of a parallel glass slab as shown in the figure below.

$Explain refraction of light through glass slab Using diagram - Science - Light - Reflection and Refraction - 11164071 | Meritnation.com$

When the light ray AE is refracted in glass, it bends towards normal, as it is optically denser than air. The ray EF, which is the refracted ray in glass is again refracted in air and bends away from the normal. It follows the path FD. Using Snell's law, we may write:

$n_{g a} = \frac{\sin i}{\sin r}$ ......(i)

$n_{a g} = \frac{\sin r^{'}}{\sin e} = \frac{\sin r}{\sin e}$ ....(ii)

Multiplying (i) and(ii), we get:

$n_{g a} \times n_{a g} = 1 = \frac{\sin i}{\sin e}$

$\Rightarrow \sin i = \sin e$

$\Rightarrow i = e$ .....(iii)

Hence, the emergent ray is parallel to the incident ray. We can see that there is some distance $(d)$ between the incident and the emergent ray. This is called lateral displacement. To find this, using trigonometric relations, we can write:

$EF = \frac{t}{cosr}$ .....(iv)

$d = EF \sin (i - r)$ .....(v)

Using(iv) and(v) e get:

$\Rightarrow d = \frac{t \sin (i - r)}{\cos r} = t \sec r \sin (i - r)$ .......(vi)

Hence, the emergent ray is parallel to the incident ray but is displaced laterally through a distance given by (vi).

Discuss refraction through a glass slab and show that emergent ray is parallel to the incident ray but displaced.

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