/*! 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 164 A plano-convex lens of refractiv... [FREE SOLUTION] | 91影视

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Chapter 18: Problem 164

A plano-convex lens of refractive index \(1.5\) and radius of curvature \(30 \mathrm{~cm}\) is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of size of the object [2004] (A) \(60 \mathrm{~cm}\) (B) \(30 \mathrm{~cm}\) (C) \(20 \mathrm{~cm}\) (D) \(80 \mathrm{~cm}\)

Short Answer

Expert verified
The object should be placed at a distance of \(60\mathrm{~cm}\) from the plano-convex lens to have a real image of the same size as the object.

Step by step solution

01

Find the focal length of the plano-convex lens

Use the lensmaker's formula to find the focal length of the plano-convex lens: \[\frac{1}{f} = (n-1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)\] where n is the refractive index of the lens, \(R_1\) is the radius of curvature of the plano side (infinity), \(R_2\) is the radius of curvature of the convex side, and f is the focal length of the lens. In this case, n = 1.5, and \(R_2 = 30 \mathrm{~cm}\). \[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{\infty} - \frac{1}{30} \right)\]
02

Calculate the focal length

Plug in the known values to find the focal length f: \[ \frac{1}{f} = 0.5 \left( -\frac{1}{30} \right) \Rightarrow f = -60\mathrm{~cm} \] The negative sign indicates that the focal length is on the opposite side of the lens from the curved surface, but the silvered surface on the convex side of the lens will reflect light back, making the effective focal length positive.
03

Apply the lens and mirror formulae

Now, we apply the lens formula and the mirror formula to determine the object distance (p) that creates an image with a size equal to the object (magnification = 1). The lens formula is: \[\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\] And the mirror formula is: \[m = \frac{-q}{p}\] Since the magnification is 1 and the lens and mirror are combined, this equation becomes: \[1 = \frac{-q}{2p}\]
04

Solve for object distance

Rearrange the equation to solve for p: \[q = -2p\] Substitute this value for q in the lens formula: \[\frac{1}{60} = \frac{1}{p} - \frac{1}{2p}\] Simplify the equation and solve for p: \[ \frac{1}{60} = \frac{1}{2p} \Rightarrow p = 2 \cdot 30 = 60\mathrm{~cm}\] Thus, the object should be placed at a distance of \(60\mathrm{~cm}\) from the lens to have a real image of the same size as the object. Therefore, the correct answer is (A).

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Most popular questions from this chapter

The plane face of plano-convex lens of focal length 20 \(\mathrm{cm}\) is silvered. This combination is equivalent to the type of mirror and its focal length is (A) convex, \(f=20 \mathrm{~cm}\) (B) concave, \(f=20 \mathrm{~cm}\) (C) convex, \(f=10 \mathrm{~cm}\) (D) concave, \(f=10 \mathrm{~cm}\)

Light wave enters from medium 1 to medium \(2 .\) Its velocity in second medium is double from first. For total internal reflection, the angle of incidence must be greater than (A) \(30^{\circ}\) (B) \(60^{\circ}\) (C) \(45^{\circ}\) (D) \(90^{\circ}\)

An object is placed in front of a convex mirror at a distance of \(50 \mathrm{~cm}\). A plane mirror is introduced covering the lower half of the convex mirror. If the distance between the object and the plane mirror is \(30 \mathrm{~cm}\), there is no parallax between the images formed by the two mirrors. The radius of curvature of the convex mirror (in \(\mathrm{cm}\) ) is (A) 60 (B) 50 (C) 30 (D) 25

In the diagram shown, the object is performing SHM according to the equation \(y=2 A \sin (\omega t)\) and the plane mirror is performing SHM according to the equation \(Y=-A \sin \left(\omega t-\frac{\pi}{3}\right)\). The diagram shows the state of the object and the mirror at time \(t=0 \mathrm{~s}\). The minimum time from \(t=0 \mathrm{~s}\) after which the velocity of the image becomes equal to zero is (A) \(\frac{\pi}{3 \omega}\) (B) \(\frac{3 \pi}{\omega}\) (C) \(\frac{\pi}{6 \omega}\) (D) \(\frac{2 \pi}{3 \omega}\)

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) \(5.0 \times 10^{-3} \mathrm{rad}\)
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