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Chapter 35: Problem 2882

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 \((\mathrm{I} / 2)\)

Short Answer

Expert verified
The correct answer is (A) \(f\) and \((3I / 4)\), as the focal length remains unchanged and the image intensity reduces to \(\frac{3}{4}\) of the original intensity I when the central part of the aperture is blocked.

Step by step solution

01

Calculate the aperture area ratio

To calculate the ratio between the areas of the obstructed and unobstructed aperture, divide the area of the obstructed aperture by the area of the original aperture. The aperture area is given by \(A = \pi r^2\), where r is the radius. For the original aperture (unobstructed), \(A_1 = \pi (\frac{d}{2})^2\) For the obstructed aperture, \(A_2 = \pi (\frac{d}{2})^2 - \pi (\frac{d}{4})^2\) The ratio between the two areas is: \(\frac{A_2}{A_1} = \frac{\pi (\frac{d}{2})^2 - \pi (\frac{d}{4})^2}{\pi (\frac{d}{2})^2} = 1 - \frac{(\frac{d}{4})^2}{(\frac{d}{2})^2} = 1 - \frac{1}{4} = \frac{3}{4}\) Next, we analyze the change in the image intensity due to this obstruction.
02

Determine the change in image intensity

Since the image intensity is directly proportional to the aperture area, the change in image intensity can be calculated as a ratio: \(\frac{I_{2}}{I_{1}} = \frac{A_2}{A_1} = \frac{3}{4}\) Now, let's evaluate the change in the lens's focal length.
03

Analyze the change in the lens's focal length

The focal length of a thin lens depends on the refractive index of the lens material and its curvature, but not its aperture size. Therefore, the focal length remains unchanged when the central part of the aperture is blocked. So, \(f_{2} = f_{1} = f\) Finally, let's compare the changes in focal length and image intensity with the given options.
04

Choose the correct answer

Comparing the changes in focal length and image intensity, we found that the new focal length remains the same as f and the image intensity reduces to \(\frac{3}{4}\) of the original intensity I. These changes correspond to the option: (A) \(f\) and \((3I / 4)\) So, the correct answer to this exercise is (A) \(f\) and \((3I / 4)\).

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Most popular questions from this chapter

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

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

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) \(\left[(\mathrm{D}-\mathrm{d})^{2} /(\mathrm{D}+\mathrm{d})^{2}\right]\)
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