If the angle of incidence is 45° then what will be the angle of refraction?


If the angle of incidence is 45° then the angle of refraction depends on the refractive index or surface material of that medium (μ).

We know that,

$\mu =\frac{\sin(i)}{\sin(r)}$

Where,

μ = refractive index of the medium,

i = angle of incident,

r = angle of refraction.

As from the above-given incident angle, 45°

$\mu =\frac{\sin45\unicode{xb0} }{\sin(r)}$

$\sin(r)=\frac{\sin(45\unicode{xb0} )}{\mu }$

$\sin(r)=\frac{1}{\sqrt{2}\mu }$ ------------ (i)

Now, from the above equation, we can conclude that the angle of refraction depends on $\mu$ of the medium through which light ray passes.

Therefore, let's assume, that light passes through a low refractive index, like air (faster medium or optically rarer medium) to a high refractive index, like water (slower medium or optically denser medium), its speed decreases and bends towards the normal. 

The refractive index of air, $\mu$ = 1

Now, putting the value of $\mu$ in the (i) we get-

$\sin(r)=\frac{1}{\sqrt{2}\times 1}=\frac{1}{\sqrt{2}}$

$\sin(r)=45\unicode{xb0} $          $[\because sin45\unicode{xb0} =\frac{1}{\sqrt{2}}]$

Thus, sin(r) = sin (i) = 45°


Let's assume, that light passes through a low refractive index, like air (faster medium or optically rarer medium) to a high refractive index, like glass (slower medium or optically denser medium), its speed decreases and bends towards the normal.

The refractive index of glass, $\mu$ = 1.5

Now, putting the value of $\mu$ in the (i) we get-

$\sin(r)=\frac{1}{\sqrt{2}\times 1.5}=\frac{1}{1.41\times 1.5}=\frac{1}{2.1}\approx \frac{1}{2}$

$\sin(r)=30\unicode{xb0} $          $[\because sin30\unicode{xb0} =\frac{1}{2}]$

Thus, sin(r) = 30°

Updated on: 10-Oct-2022

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