In the given figure rays incident on an interface would converge 2 cm below the interface if they continued to move in straight lines without bending. But due to refraction, the rays will bend and mee

Question

In the given figure rays incident on an interface would converge 2 cm below the interface if they continued to move in straight lines without bending. But due to refraction, the rays will bend and meet somewhere else. Find the distance of meeting point of refracted rays below the interface (in cm), assuming the rays to be making small angles with the normal to the interface.

TestHub · Accepted Answer

Let n 1 ​ be the refractive index of air (approximately 1) and n 2 ​ = 2.5 be the refractive index of the medium. Let u = − 2 cm be the object distance. Using the formula for refraction at a spherical surface (in this case, a flat surface), we have: v n 2 ​ ​ − u n 1 ​ ​ = R n 2 ​ − n 1 ​ ​ Since the interface is flat, R = ∞ , so R n 2 ​ − n 1 ​ ​ = 0 . Thus, v n 2 ​ ​ = u n 1 ​ ​ v = n 1 ​ n 2 ​ ​ u = 1 2.5 ​ ( − 2 ) = − 5 c m The rays will meet 5 cm below the interface.

Physics - Geometrical Optics Question with Solution | TestHub

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