NEET Refraction by Prism MCQs & PYQs

Practice NEET Refraction by Prism Questions

Question 11:

easy

The refractive index of the material of a prism is $\sqrt{2}$ and its refracting angle is $30^\circ$. One of the refracting surfaces of the prism is made a mirror inwards. A beam of monochromatic light entering the prism from the other face will retrace its path after reflection from the mirrored surface if its angle of incidence on the prism is:

(2004)

For the beam to retrace its path, it must hit the mirrored surface normally, so $r_2 = 0$. Since $r_1 + r_2 = A$, we have $r_1 = 30^\circ$. Using Snell's law, $1 \cdot \sin(i) = \mu \sin(r_1) = \sqrt{2} \sin(30^\circ) = \sqrt{2}(1/2) = 1/\sqrt{2}$. Therefore, the angle of incidence is $i = 45^\circ$.

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Question 12:

easy

For a prism its refractive index is $\cot A/2$ then minimum angle of deviation is:

(1999)

Refractive index $\mu = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)}$. Given $\mu = \cot(A/2) = \frac{\cos(A/2)}{\sin(A/2)} = \frac{\sin(90^\circ - A/2)}{\sin(A/2)}$. Equating numerators: $\frac{A + \delta_m}{2} = 90^\circ - \frac{A}{2}$. This gives $A + \delta_m = 180^\circ - A$, so the minimum angle of deviation is $\delta_m = 180^\circ - 2A$.

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Question 13:

easy

There is a prism with refractive index equal to $\sqrt{2}$ and the refractive angle equal to $30^\circ$. One of the refractive surface of the prism is polished. A beam of monochromatic light will retrace its path if its angle of incidence over the refracting surface of the prism is

(1992)

For the ray to retrace its path, it must strike the silvered surface normally, so $r_2 = 0$. From $r_1 + r_2 = A$, we have $r_1 = 30^\circ$. Using Snell's law at the first surface, $1 \cdot \sin(i) = \mu \sin(r_1) = \sqrt{2} \sin(30^\circ) = \sqrt{2}(1/2) = 1/\sqrt{2}$. Therefore, the angle of incidence is $i = 45^\circ$.

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Question 14:

easy

A ray is incident at an angle of incidence i on one surface of a prism of small angle A and emerge normally from opposite surface. If the refractive index of the material of prism is $\mu$, the angle of incidence i is nearly equal to:

(1989)

Since the ray emerges normally, the angle of emergence $e = 0$, which implies $r_2 = 0$. For a prism, $r_1 + r_2 = A$, so $r_1 = A$. Using Snell's law for small angles, $i = \mu r_1$, which gives the angle of incidence $i = \mu A$.

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Question 15:

easy

Pick the wrong answer in the context with rainbow.

(2019)

An observer can only see a rainbow when looking away from the sun, so the sun must be behind them. Thus, the statement that the observer's front is towards the sun is incorrect.

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