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Chapter 35: Q. 6 (page 1017)

A $2.0 m$ tall man is $10 m$ in front of a camera with a $15 mm$ focal length lens. How tall is his image on the detector?

Short Answer

Expert verified

The height of the image will be $0.0015 m .$

Step by step solution

01

Given information

We have given,

Height of the man = $2 m$

Distance from the camera = $10 m$

focal length of the camera = $15 mm = 0.015 m$

We have to find the height of the image.

02

Simplify

Since the camera uses a converging lens.

then using the lens's equation

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \frac{1}{0.015} = \frac{1}{v} - \frac{1}{- 10} \frac{1}{v} = \frac{1}{0.015} - \frac{1}{10} = \frac{9.985}{0.15} v = 0.015 m$

Since the image formed by the lens is inverted.

$m = \frac{H e i g h t o f t h e i m a g e}{H e i g h t o f t h e o b j e c t} = \frac{h'}{h} \frac{h'}{2} = \frac{v}{u} = \frac{0.015}{10} h' = - 0.0015 m$

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Most popular questions from this chapter

Suppose you wanted special glasses designed to let you see underwater without a face mask. Should the glasses use a converging or diverging lens? Explain.

In figure $p 35.26$ , parallel rays from the left are focused to a point at two locations on the optical axis. find the position of each location, giving your answer as a distance left or right of the lens.

A friend lends you the eyepiece of his microscope to use on your own microscope. He claims the spatial resolution of your microscope will be halved, since his eyepiece has the same diameter as yours but twice the magnification. Is his claim valid? Explain.

A magnifier has a magnification of $5 x$ . How far from the lens should an object be held so that its image is seen at the near-point distance of $25 c m$ ? Assume that your is immediately behind the lens.

FIGURE $C P 35.50$ shows a simple zoom lens in which the magnitudes of both focal lengths are $f$ . If the spacing $d < f$ , the image of the converging lens falls on the right side of the diverging lens. Our procedure of letting the image of the first lens act as the object of the second lens will continue to work in this case if we use a negative object distance for the second lens. This is called a virtual object. Consider an object very far to the left $(s \approx \infty)$ of the converging lens. Define the effective focal length as the distance from the midpoint between the lenses to the final image.

a. Show that the effective focal length is

$f_{e f f} = \frac{f^{2} - f d + \frac{1}{2} d^{2}}{d}$

b. What is the zoom for a lens that can be adjusted from $d = \frac{1}{2} f$ to $d = \frac{1}{4} f$ ?

See all solutions

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