A ray of light passing through a prism having refractive index √2 suff...
Solution:
Given:
Refractive index of prism, $\mu = \sqrt{2}$
Angle of incidence, $i = 2r$
Let us first draw a diagram of the given situation:
![image.png](attachment:image.png)
Minimum Deviation:
For a given prism, the angle of incidence and angle of emergence for which the deviation produced is minimum is called the angle of minimum deviation. Let this angle be $D_m$.
We know that for a prism, angle of deviation is given by:
$$A = (μ − 1) (i + e − A)$$
Where, $i$ is the angle of incidence, $e$ is the angle of emergence and $A$ is the angle of prism.
For minimum deviation, $\frac{dA}{de} = 0$
Differentiating the above equation with respect to $e$, we get:
$$\frac{dA}{de} = \frac{(μ-1)}{2} \left(\frac{\sin i}{\sin(\frac{A+e}{2})} - \frac{\sin e}{\sin(\frac{A+e}{2})} - \cos(\frac{A+e}{2})\right) = 0$$
Simplifying the above equation, we get:
$$\sin i = (μ − 1) \sin \left(\frac{A}{2}\right)$$
Angle of Deviation:
Using Snell's Law, we can write:
$$\frac{\sin i}{\sin r} = \mu$$
As given, $i = 2r$
Substituting this value in the above equation, we get:
$$\frac{\sin 2r}{\sin r} = \sqrt{2}$$
Simplifying the above equation, we get:
$$2\cos r = \sqrt{2}$$
$$\cos r = \frac{\sqrt{2}}{2}$$
$$r = 45º$$
Therefore, $i = 90º$
Angle of Prism:
Using the relation $A = (μ − 1) (i + e − A)$, we can write:
$$A = (μ − 1) (i + e -A)$$
Substituting the values of $μ$, $i$ and $e$ in the above equation, we get:
$$A = (\sqrt{2}-1)(90º+45º-A)$$
Simplifying the above equation, we get:
$$A = 60º$$
Therefore, the angle of prism is $60º$.
A ray of light passing through a prism having refractive index √2 suff...
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