Published by:
CGP EDU Academic Team
Published on: September 12, 2026
A ray of light undergoes deviation of 30º when incident on an equilateral prism of refractive index
. The angle made by the ray inside the prism with the base of the prism is ...........
Text Solution
Verified by ExpertsThe correct answer is:
A
Step 1: Given that the prism is equilateral, the angle of the prism \( A = 60^\circ \).
Step 2: The angle of deviation \( D = 30^\circ \).
Step 3: The relation for deviation through a prism is given by:
\( D = A + (n - 1) \cdot A \) where \( n \) is the refractive index.
Step 4: We know \( D = 30^\circ \) and \( A = 60^\circ \).
Step 5: Rearranging gives us: \( n = \frac{D}{A - D} = \frac{30}{60 - 30} = \frac{30}{30} = 1 \).
Step 6: To find the internal angle \( \theta \) made with the base of the prism, we use the formula for internal refraction: \( n = \frac{\sin(i)}{\sin(r)} \). Since all calculations are approximations, we will focus on the geometric layout of the prism.
Step 7: The angle \( r\) inside the prism can be approximated too: \(\theta = \frac{A}{2} + \frac{D}{2} = \frac{60^\circ}{2} + \frac{30^\circ}{2} = 30^\circ + 15^\circ = 45^\circ\).
Therefore, the angle made by the ray inside the prism with the base of the prism is approximately \( 45^\circ \).
Step 2: The angle of deviation \( D = 30^\circ \).
Step 3: The relation for deviation through a prism is given by:
\( D = A + (n - 1) \cdot A \) where \( n \) is the refractive index.
Step 4: We know \( D = 30^\circ \) and \( A = 60^\circ \).
Step 5: Rearranging gives us: \( n = \frac{D}{A - D} = \frac{30}{60 - 30} = \frac{30}{30} = 1 \).
Step 6: To find the internal angle \( \theta \) made with the base of the prism, we use the formula for internal refraction: \( n = \frac{\sin(i)}{\sin(r)} \). Since all calculations are approximations, we will focus on the geometric layout of the prism.
Step 7: The angle \( r\) inside the prism can be approximated too: \(\theta = \frac{A}{2} + \frac{D}{2} = \frac{60^\circ}{2} + \frac{30^\circ}{2} = 30^\circ + 15^\circ = 45^\circ\).
Therefore, the angle made by the ray inside the prism with the base of the prism is approximately \( 45^\circ \).
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