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Chapter 31: Problem 47

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) \(f-10 \mathrm{cm} .\)

Short Answer

Expert verified
To summarize, calculate image distance for the given object distances using the lens formula, and then find the height and type of image using magnification and sign of image distance respectively. Repeat for different object distances. Details would depend upon actual calculations.

Step by step solution

01

Calculate Image Distance for \( f + 10 \) cm

Given that \( f = 35 \) cm and \( u = -(f + 10) \) cm (negative because object is always placed to the left of the lens), use the lens formula \( 1/f = 1/v - 1/u \) to find \( v \).
02

Determine Type and Height of Image for \( f + 10 \) cm

The type of the image depends on the sign of \( v \). If \( v \) is positive, image is real and inverted. If \( v \) is negative, image is virtual and erect. Use the magnification formula \( h'_i/h_o = -v/u \) to find the height of the image.
03

Repeat Calculations for \( f - 10 \) cm

Repeat steps 1 and 2 using \( u = -(f - 10) \) cm to find \( v \) and the height of the image.

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Most popular questions from this chapter

A candle and a screen are \(70 \mathrm{cm}\) apart. Find two points between candle and screen where you could put a convex lens with \(17-\mathrm{cm}\) focal length to give a sharp image of the candle on the screen.

At what two distances could you place an object from a \(45-\mathrm{cm}-\) focal-length concave mirror to get an image 1.5 times the object's size?

An object \(15 \mathrm{cm}\) from a concave mirror has a virtual image magnified 2.5 times. What's the mirror's focal length?

A 12 -mm-high object is \(10 \mathrm{cm}\) from a concave mirror with focal length \(17 \mathrm{cm} .\) (a) Where is the image, (b) how high is it, and (c) what type is it?

A diverging lens always makes a reduced image. Could you use such a lens to start a fire by focusing sunlight? Explain.
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