/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 183 A thin convex lens made from cro... [FREE SOLUTION] | 91影视

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Chapter 18: Problem 183

A thin convex lens made from crown glass \(\left(\mu=\frac{3}{2}\right)\) has focal length \(f\). When it is measured in two different liquids having refractive indices \(\frac{4}{3}\) and \(\frac{5}{3}\), it has the focal lengths \(f_{1}\) and \(f_{2}\), respectively. The correct relation between the focal lengths is (A) \(f_{1}=f_{2}f\) and \(f_{2}\) becomes negative (C) \(f_{2}>f\) and \(f_{1}\) becomes negative (D) \(f_{1}\) and \(f_{2}\) both become negative

Short Answer

Expert verified
Based on the computation, the correct answer is (D) - \(f_{1}\) and \(f_{2}\) both become negative when the lens is immersed in both liquids.

Step by step solution

01

Lens in Air

Initially, the lens is in air, and the refractive index of air is 1. So, applying the lens-maker's formula, the focal length of the lens is given as \(\tfrac{1}{f} = \tfrac{\mu - 1}{1} \times (\tfrac{1}{R_1}-\tfrac{1}{R_2})\).
02

Lens in First Liquid

When the lens is placed in the first liquid with a refractive index of \(\tfrac{4}{3}\), the formula changes to \(\tfrac{1}{f_{1}} = \tfrac{\tfrac{3}{2} - \tfrac{4}{3}}{\tfrac{4}{3}} \times (\tfrac{1}{R_1} - \tfrac{1}{R_2})\). This simplifies to \(\tfrac{1}{f_{1}} = \tfrac{-2}{3} \times (\tfrac{1}{R_1} - \tfrac{1}{R_2})\). Seeing that the right-hand side is negative, which implies \(f_{1}<0\). Therefore, the focal length becomes negative in the first liquid.
03

Lens in Second Liquid

When the lens is placed in the second liquid with refractive index of \(\tfrac{5}{3}\), the formula changes to \(\tfrac{1}{f_{2}} = \tfrac{\tfrac{3}{2} - \tfrac{5}{3}}{\tfrac{5}{3}} \times (\tfrac{1}{R_1} - \tfrac{1}{R_2})\). This simplifies to \(\tfrac{1}{f_{2}} = \tfrac{-1}{5} \times (\tfrac{1}{R_1} - \tfrac{1}{R_2})\), which also implies that \(f_{2}<0\). Therefore, the focal length becomes negative in the second liquid.

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Most popular questions from this chapter

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