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Chapter 35: Problem 2888

An object is placed at a distance of \((\mathrm{f} / 2)\) the from a convex lens the image will be.... (A) at \(\mathrm{f}\), real and inverted (B) at,\((3 \mathrm{f} / 2)\) real and inverted (C) at one of the foci, virtual and double its size (D) at \(2 \mathrm{f}\), virtual and erect.

Short Answer

Expert verified
The correct answer is (D) at \(2 \mathrm{f}\), virtual and erect.

Step by step solution

01

Write down the given information and the lens formula

The object is at a distance of \((\mathrm{f}/2)\) from the convex lens. The lens formula we will use is: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\] where \( f\) is the focal length of the convex lens, \(d_o\) is the object distance, and \(d_i\) is the image distance.
02

Use the given information to write the equation

The object distance, in this case, is \(d_o = \frac{f}{2}\). Plug this into the lens formula: \[\frac{1}{f} = \frac{1}{\frac{f}{2}} + \frac{1}{d_i}\]
03

Solve the equation for the image distance

To solve the equation, we need to eliminate the fractions. Multiply each term by \(2f\) to simplify: \[2 = \frac{2f}{f} + \frac{2f}{d_i}\] \[2 = 2 + \frac{2f}{d_i}\] Now, solve for \(d_i\): \[\frac{2f}{d_i} = 0\] It seems like we have reached a dead end due to the result \(2f = 0\), but we should note that this result is not possible because the focal length cannot be zero. Due to the assumptions made during the derivation of the lens formula, it is not valid for distances close to the focal length. However, we know that when an object is close to the focal length of a lens, a virtual and erect image is formed. We can now determine the nature of the image based on this information.
04

Determine if the image is real or virtual, erect, or inverted based on the approximate object distance

Since the object distance is close to the focal length, we know the image formed will be virtual and erect. The only option that satisfies these conditions is option (D).
05

Confirm the answer

Thus, the correct answer, given the limits of the lens formula, is: (D) at \(2 \mathrm{f}\), virtual and erect.

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Most popular questions from this chapter

A convex lens of focal length \(\mathrm{f}\) is placed somewhere in between an object and a screen. The distance between the object and the screen is \(\mathrm{x}\). If the numerical value of the magnification product by the lens is \(\mathrm{m}\), What is the focal length of the lens? (A) \(\left[\mathrm{mx} /(\mathrm{m}-1)^{2}\right]\) (B) \(\left[\mathrm{mx} /(\mathrm{m}+1)^{2}\right]\) (C) \(\left[(m-1)^{2} / \mathrm{m}\right] \mathrm{x}\) (D) \(\left[(\mathrm{m}+1)^{2} / \mathrm{m}\right] \mathrm{x}\)

A spherical mirror forms an erect image three times the linear size of the object. If the distance between the object and the image is \(80 \mathrm{~cm}\), What is the focal length of the mirror? (A) \(30 \mathrm{~cm}\) (B) \(40 \mathrm{~cm}\) (C) \(-15 \mathrm{~cm}\) (D) \(15 \mathrm{~cm}\)

A short linear object of length \(L\) lies on the axis of a spherical mirror of focal length of \(f\) at a distance \(u\) from the mirror. Its image has an axial length \(L^{\prime}\) equal to \(\ldots \ldots \ldots\).. (A) \(\mathrm{L}[\mathrm{f} /(\mathrm{u}-\mathrm{f})]^{2}\) (B) \(\mathrm{L}[(\mathrm{u}-\mathrm{f}) / \mathrm{f}]^{2}\) (C) \(\mathrm{L}[(\mathrm{u}+\mathrm{f}) / \mathrm{f}]^{1 / 2}\) (D) \(L[f /(u-f)]^{1 / 2}\)

A concave mirror of focal length \(\mathrm{f}\) produces an images n times the size of the object. If the image is real then What is the distance of the object from the mirror? (A) \((\mathrm{n}+1) \mathrm{f}\) (B) \([(\mathrm{n}-1) / \mathrm{n}] \mathrm{f}\) (C) \((\mathrm{n}-1) \mathrm{f}\) (D) \([(\mathrm{n}+1) / \mathrm{n}] \mathrm{f}\)

A concave lens of focal length \(\mathrm{f}\) forms an image which is n times the size of the object. What is the distance of the object from the lens? (A) \((1+\mathrm{n}) \mathrm{f}\) (B) \((1-\mathrm{n}) \mathrm{f}\) (C) \([(1-\mathrm{n}) / \mathrm{n}] \mathrm{f}\) (D) \([(1+\mathrm{n}) / \mathrm{n}] \mathrm{f}\)
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