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Chapter 31: Problem 77

The maximum magnification of a simple magnifier occurs with the image at the 25 -cm near point. Show that the angular magnification is \(m=1+(25 \mathrm{cm} / \mathrm{f}),\) where \(f\) is the focal length.

Short Answer

Expert verified
The formula to calculate the maximum angular magnification of a simple magnifier when the image is at the near point of 25cm is provided by \(m = 1 + \frac{25 cm}{f}\), where \(f\) is the focal length of the magnifier.

Step by step solution

01

Understanding Angular Magnification

Angular magnification tells about the ratio of the angle subtended by the image to the angle subtended by the object at the eye. In this case, the angular magnification \(m\) is given as \(m = 1 + \frac{d}{f}\) where \(d\) is the distance from the eye to the image.
02

Considering the Near Point

The near point for a standard or average human eye is typically at 25cm. This is the closest distance an object can be to the eye and still be in clear focus. Thus let's substitute the value of \(d\) with 25cm in our formula.
03

Finalizing the Expression

Replacing the value of \(d\) in the formula: \(m = 1 + \frac{25 cm}{f}\) we get our expression for the angular magnification of a simple magnifier where the image is at the 25cm near point.

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Most popular questions from this chapter

A double-convex lens with equal 28.5 -cm curvature radii is made from glass with refractive indices \(n_{\mathrm{red}}=1.512\) and \(n_{\text {violet }}=1.547 .\) If a point source of white light is located on the lens axis at \(75.0 \mathrm{cm}\) from the lens, over what distance will its visible image be smeared?

A lens has focal length \(f=35 \mathrm{cm} .\) Find the type and height of the image produced when a 2.2 -cm-high object is placed at distances (a) \(f+10 \mathrm{cm}\) and (b) \(f-10 \mathrm{cm} .\)

A candle is on the axis of a \(15-\mathrm{cm}\) -focal-length concave mirror, \(36 \mathrm{cm}\) from the mirror. (a) Where is its image? (b) How do the image and object sizes compare? (c) Is the image real or virtual?

If you're handed a converging lens, what can you do to estimate its focal length quickly?

A lightbulb is \(56 \mathrm{cm}\) from a convex lens. Its image appears on a screen \(31 \mathrm{cm}\) from the lens, on the other side. Find (a) the lens's focal length and (b) how much the image is enlarged or reduced.
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