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Chapter 35: Problem 2883

The distance between object and the screen is D. Real images of an object are formed on the screen two positions of a lens separated by a distance \(\mathrm{d}\). What will be the ratio between the sizes of two images? (A) \(\left(\mathrm{D}^{2} / \mathrm{d}^{2}\right)\) (B) \((\mathrm{D} / \mathrm{d})\) (C) \(\sqrt{(\mathrm{D} / \mathrm{d})}\) (D) \(\left[(\mathrm{D}-\mathrm{d})^{2} /(\mathrm{D}+\mathrm{d})^{2}\right]\)

Short Answer

Expert verified
The correct answer for the ratio between the sizes of the two images is \(\frac{h_{1}}{h_{2}} = \frac{v_{1}}{v_{2}} = \frac{d}{(D - p)(D - (p-d))}\), which is not found among the given options (A), (B), (C), or (D).

Step by step solution

01

Understand the lens formula.

The lens formula is given by \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\), where \(f\) is the focal length of the lens, \(u\) is the object distance, and \(v\) is the image distance.
02

Find the image size for both positions of the lens.

Since the object distance is fixed at D, the image size is proportional to the image distance (\(\frac{h_{1}}{h_{2}} = \frac{v_{1}}{v_{2}}\)). Let the two positions of the lens be \(p\) and \(q\). Then we have: For position \(p\), object distance \(u_{1} = D − p\), image distance \(v_{1}\) For position \(q\), object distance \(u_{2} = D − q\), image distance \(v_{2}\)
03

Solve for image distances using the lens formula.

Using the lens formula for positions \(p\) and \(q\), we get: For position \(p\), \(\frac{1}{f} = \frac{1}{D-p} + \frac{1}{v_{1}}\) For position \(q\), \(\frac{1}{f} = \frac{1}{D-q} + \frac{1}{v_{2}}\) Subtract the second equation from the first: \(\frac{1}{v_{1}} - \frac{1}{v_{2}} = \frac{1}{D-p} - \frac{1}{D-q}\)
04

Find the image size ratio.

Now, find the ratio \(\frac{h_{1}}{h_{2}} = \frac{v_{1}}{v_{2}}\). From the equation in Step 3, rearrange terms to get: \(\frac{1}{v_{1}} - \frac{1}{v_{2}} = \frac{q-p}{\left(D - p\right)\left(D - q\right)}\) Combining the terms, we get: \(\frac{v_{1} - v_{2}}{v_{1}v_{2}} = \frac{q-p}{\left(D - p\right)\left(D - q\right)}\) Hence, the ratio between the sizes of the two images is: \(\frac{h_{1}}{h_{2}} = \frac{v_{1}}{v_{2}} = \frac{q-p}{\left(D - p\right)\left(D - q\right)}\)
05

Express the image size ratio in terms of D and d.

We have \(d = |p - q|\). If we rearrange and insert as \((p - q) = d\), we get: \(\frac{h_{1}}{h_{2}} = \frac{d}{\left(D - p\right)\left(D - (p-d)\right)} = \frac{d}{\left(D - p\right)\left(D - (p-d)\right)}\) This ratio cannot be simplified further into any of the given options (A), (B), (C), or (D). Therefore, it appears that a correct answer for the ratio between the sizes of the two images is not given among the provided options.

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Most popular questions from this chapter

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) \([(1+\mathrm{n}) / \mathrm{n}] \mathrm{f}\)

A thin lens has focal length \(\mathrm{f}\), and its aperture has diameter d. It forms an image of intensity I. Now, the central part of the aperture upto diameter \((\mathrm{d} / 2)\) is blocked by an opaque paper. The focal length and image intensity will change to \(\ldots\) (A) \(\mathrm{f}\) and \((3 \mathrm{I} / 4)\) (B) \((3 \mathrm{f} / 4)\) and \((\mathrm{I} / 2)\) (C) \(\mathrm{f}\) and \((\mathrm{I} / 4)\) (D) \((\mathrm{f} / 2)\) and \((\mathrm{I} / 2)\)

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 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 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) \(15 \mathrm{~cm}\)
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