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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°