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 :
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:
- 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 \(30^\circ\).
- This situation is depicted by using Snell’s Law at the first surface: \(n \sin i = \sin r\), where \(i = 50^\circ\) and \(r = 30^\circ\).
- 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.
- Since \(r_1 = r_2\) and the angle of the prism \(A = 60^\circ\), the angle of emergence \(e\) is also \(50^\circ\).
- 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.
Similar Questions
- In interference and diffraction, the light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe.
A. As there is no gain or loss of energy, these phenomena are consistent with the principle of conservation of energy.
B. Diffraction and interference are characteristics exhibited only by light waves.
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The lens combination as shown in the figure, consists of two lenses, $L_1$ and $L_2$, of the focal lengths $+10\text{ cm}$ and $-10\text{ cm}$, respectively. The position of the image formed is :
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