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Question

A ray of monochromatic light is passing through an equilateral prism (ABC) as shown in the figure. The refracted ray (QR) is parallel to the base (BC) and the angle of incidence ($i$) is $50^\circ$. Then the angle of deviation ($\delta$) is :

This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
$40^\circ$

To solve this problem, it is essential to understand the refraction of light through a prism and how the angle of deviation is determined in such scenarios.

Given Data:

  • The prism is equilateral, meaning all its angles are \(60^\circ\).
  • The angle of incidence \(i = 50^\circ\).
  • The refracted ray \(QR\) is parallel to the base \(BC\).

Concept and Calculation:

  1. In an equilateral prism, the refracted ray \(QR\) being parallel to the base indicates that the angle of refraction at \(Q\) and exit angle at \(R\) combined will equal the angle of the prism. Therefore, the angle of refraction \(r\) at both \(Q\) and \(R\) is \(30^\circ\).
  2. This situation is depicted by using Snell’s Law at the first surface: \(n \sin i = \sin r\), where \(i = 50^\circ\) and \(r = 30^\circ\).
  3. The angle of deviation \(\delta\) is given by the formula: \(\delta = i + e - A\), where \(e\) is the angle of emergence and \(A\) is the angle of the prism.
  4. Since \(r_1 = r_2\) and the angle of the prism \(A = 60^\circ\), the angle of emergence \(e\) is also \(50^\circ\).
  5. Substituting the values, we have: \(\delta = 50^\circ + 50^\circ - 60^\circ = 40^\circ\).

Therefore, the angle of deviation \(\delta\) is 40°.

Conclusion: The correct answer is 40°, as it matches the angle of deviation calculated.

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Important Questions from Optics

  1. In a microscope the objective is having focal length $f_o = 2 \text{ cm}$ and eye-piece is having focal length $f_e = 4 \text{ cm}$. The tube length is $32 \text{ cm}$. The magnification produced by this microscope for normal adjustment is ____________.
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