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(C) The deviation angle $\delta$ produced by a thin prism is given by the formula $\delta = A(\mu - 1)$.Given $A = 5^{\circ} = 5 	imes \frac{\pi}{180

Vedclass · Accepted Answer

$\frac{5\pi}{36}\,cm$

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(C) The deviation angle $\delta$ produced by a thin prism is given by the formula $\delta = A(\mu - 1)$.Given $A = 5^{\circ} = 5 	imes \frac{\pi}{180}$ radians and $\mu = 1.5$.Substituting the values: $\delta = (5 	imes \frac{\pi}{180}) 	imes (1.5 - 1) = \frac{5\pi}{180} 	imes 0.5 = \frac{2.5\pi}{180} = \frac{5\pi}{360}$ radians.The lateral displacement or the shift of the image formed by a thin prism for an object at distance $d$ is given by $y = d 	imes \delta$ (for small angles).Here,$d = 10\,cm$.Therefore,the distance of the image from the object is $y = 10 	imes \frac{5\pi}{360} = \frac{50\pi}{360} = \frac{5\pi}{36}\,cm$.

$A$ thin prism of angle $5^{\circ}$ is placed at a distance of $10\,cm$ from an object. What is the distance of the image from the object? (Given refractive index $\mu$ of prism $= 1.5$)

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