/*! 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 65 A plano-convex lens has a thickn... [FREE SOLUTION] | 91影视

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Chapter 18: Problem 65

A plano-convex lens has a thickness of \(4 \mathrm{~cm}\). When placed on a horizontal table with curved surface in contact with it, the apparent depth of the bottom-most point of the lens is found to be \(3 \mathrm{~cm}\). If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of plane face is found to be \(25 / 8 \mathrm{~cm}\). The focal length of the lens is (A) \(50 \mathrm{~cm}\) (B) \(75 \mathrm{~cm}\) (C) \(100 \mathrm{~cm}\) (D) \(150 \mathrm{~cm}\)

Short Answer

Expert verified
The focal length of the lens is \(50 \mathrm{~cm}\) (A).

Step by step solution

01

Recall the formula for apparent depth

First, let's recall the formula for apparent depth involving the real depth, refractive index, and apparent depth: \(t_a = \frac{t_r}{n}\), where \(t_a\) is the apparent depth, \(t_r\) is the real depth and \(n\) is the refractive index of the material.
02

Use the given apparent depths to set up equations

Now we have enough information to set up two equations involving the refractive index: In the first position (curved surface in contact with the table), we have the following apparent depth and lens thickness: \(3 = \frac{4}{n}\) (1) In the second position (plane face in contact with the table), we have the following apparent depth: \(\frac{25}{8} = \frac{4}{n-1}\) (2)
03

Solve for the refractive index

Next, we will solve the system of equations (1) and (2) to find the refractive index. Divide equation (1) by 3 and equation (2) by \(\frac{25}{8}\): \(\frac{4}{3n} = 1\) (3) \(\frac{8}{25n - 25} = 1\) (4) Now let's solve equation (3) for \(n\): \(n = \frac{4}{3}\) (5) Substitute the value of \(n\) from equation (5) into equation (4) to find the value of \(n\): \(\frac{8}{25 (\frac{4}{3}) - 25} = 1\) Now solve for \(n\): \(n = \frac{3}{2}\)
04

Apply the lens maker's formula

The lens maker's formula is given by: \(\frac{1}{f} = (n - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right)\) Since it's a plano-convex lens, one of the radii (\(R_2\)) is infinite. In this case, the lens maker's formula becomes: \(\frac{1}{f} = (n - 1)\left( \frac{1}{R_1} \right)\) Now we know the refractive index (\(n = \frac{3}{2}\)). To find the focal length, we need to determine the radius of curvature of the lens (\(R_1\)). Using the apparent depth formula with the second position information: \(\frac{25}{8} = \frac{4}{\frac{1}{2}}\) Solve for \(R_1\): \(R_1 = 8\)
05

Calculate the focal length of the lens

Now we can plug the values of \(n\) and \(R_1\) into the lens maker's formula to find the focal length (\(f\)): \(\frac{1}{f} = (\frac{3}{2} - 1)\left( \frac{1}{8} \right)\) Solve for \(f\): \(f = 50 cm\) So, the focal length of the lens is 50 cm, which corresponds to the answer (A).

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Most popular questions from this chapter

In an interference arrangement, similar to Young's double slit experiment, the slits \(S_{1}\) and \(S_{2}\) are illuminated with coherent microwave sources, each of frequency \(10^{6} \mathrm{~Hz}\). The sources are synchronized to have zero phase difference. The slits are separated by distance \(d=150.0 \mathrm{~m}\). The intensity \(I_{(\theta)}\) is measured as a function of \(\theta\), where \(\theta\) is defined as shown in the Fig. 18.53. If \(I_{0}\) is maximum intensity, then \(I_{(\theta)}\) for \(0 \leq \theta \leq 90\) is given by (A) \(I_{(\theta)}=I_{0}\) for \(\theta=0^{\circ}\) (B) \(I_{(\theta)}=\left(I_{0} / 2\right)\) for \(\theta=30^{\circ}\) (C) \(I_{(\theta)}=\left(I_{0} / 4\right)\) for \(\theta=90^{\circ}\) (D) \(I_{(\theta)}\) is constant for all values of \(\theta\)

A fish looks up through the water sees the outside world contained in a circular horizon. If the refractive index of water is \(4 / 3\) and the fish is \(12 \mathrm{~cm}\) below the surface, the radius of this circle in \(\mathrm{cm}\) is \(\quad\) [2005] (A) \(\frac{36}{\sqrt{7}}\) (B) \(36 \sqrt{7}\) (C) \(4 \sqrt{5}\) (D) \(36 \sqrt{5}\)

A telescope of diameter \(2 \mathrm{~m}\) uses light of wavelength \(5000 \AA\) for viewing stars. The minimum angular separation between two stars whose image is just resolved by this telescope is (A) \(4 \times 10^{-4} \mathrm{rad}\) (B) \(0.25 \times 10^{-6} \mathrm{rad}\) (C) \(0.31 \times 10^{-6} \mathrm{rad}\) (D) \(5.0 \times 10^{-3} \mathrm{rad}\)

In Young's double slit experiment, double slit of separation \(0.1 \mathrm{~cm}\) is illuminated by white light. A coloured interference pattern is formed on a screen \(100 \mathrm{~cm}\) away. If a pinhole is located on this screen at a distance of \(2 \mathrm{~mm}\) from the central fringe, the wavelength in the visible spectrum which will be absent in the light transmitted through the pin hole are (A) \(5714 \AA\) and \(4444 \AA\) (B) \(6000 \AA\) and \(5000 \AA\) (C) \(5500 \AA\) and \(4500 \AA\) (D) \(5200 \AA\) and \(4200 \AA\)

In an usual Young's double slit experiment with \(\lambda=5000 \AA, d=2 \mathrm{~mm}\), and \(D=1 \mathrm{~m}\), the number of maximum obtained on the screen if it is large enough to receive any number of fringes (A) 8001 (B) 6001 (C) 4001 (D) 2001
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