Angle of prism is A and its one surface is silvered. Light rays falling at an angle of incidence 2A on first surface return back through the same path after suffering reflection at second silvered surface. Refractive index of the material of prism is?

Question

Answer

Explanation:

The angle of incidence on the first surface of the prism is 2A.

After entering the prism, the light ray undergoes refraction and moves towards the second surface.

On reaching the second surface, the light ray is reflected back and returns through the same path.

Since the angle of incidence and angle of reflection are equal, the angle of incidence on the second surface is also 2A.

Let the angle of deviation be D.

Therefore, the total deviation of the light ray is 2D.

As per the laws of refraction and reflection, the angle of deviation can be expressed as:

$D=\frac{(A-r)+(A-2r+ heta)+(A-r)}{2}-A$

Where r is the angle of refraction at the first surface and θ is the angle of reflection at the second surface.

On simplifying the above expression, we get:

$D=2r- heta$

As the light ray returns back through the same path, the deviation is zero.

Therefore, we have:

$2r- heta=0$

On solving the above equation, we get:

$heta=2r$

Substituting the value of θ in the expression for D, we get:

$D=0$

As the light ray undergoes no deviation, the prism must be symmetric.

Therefore, the angle of prism A is equal to 45 degrees.

Using the formula for the refractive index of a prism, we get:

$n=\frac{\sin(\frac{A+r}{2})}{\sin(\frac{A}{2})}$

Substituting the value of A as 45 degrees, we get:

$\large n=\frac{\sin(\frac{45+r}{2})}{\sin(\frac{45 src=$

Answer

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