Calculating the Angle Between Virtual Sources of Fresnel's Biprism

[Shovon00000]

Why is the angle in radians when we calculate the distance between the virtual sources of Fresnel's biprism??

 

[tech99]

It seems to be an approximation for the tan of the very small angle α.

 

[Steve4Physics]

Why is the angle in radians when we calculate the distance between the virtual sources of Fresnel's biprism??

The formula (using your symbols) for the separation between the sources is really:
$d = 2(\mu – 1) sin(\alpha) y_1$.

But since $\alpha$ is small, we can use the small angle approximation. There is very little difference between the values of $sin(\alpha)$ and $\alpha$ in radians.

$2º = 2 \times \frac {\pi}{180}$ radians $= \frac {\pi}{90}$ radians $= 0.0349$ radians
$sin(2º) = 0.0349$
(Check it for yourself on a calculator.)

So we simplify the above formula to:
$d = 2(\mu – 1) \alpha y_1$.

By the way, the working in your attachment has a mistake. It says:
$d = \frac {2(1.5 – 1) \times 10} {90} = ...$
but it should say:
$d = \frac {2(1.5 – 1) \times \pi \times 10} {90} = ...$