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Chapter 31: Problem 51

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 -cm focal length to give a sharp image of the candle on the screen.

Short Answer

Expert verified
The two points between the candle and screen where the convex lens can be placed are approximately 25.1 cm and 44.9 cm from the candle.

Step by step solution

01

Identify the Given Values

The total distance between the object (candle) and image (screen) is given as \(70 \mathrm{cm}\) and the focal length (\(f\)) of the convex lens is given as \(17 \mathrm{cm}\). Also, we can consider that the convex lens forms real images which means that the image distance (\(v\)) is positive.
02

Set Up the Lens Equation for the First Point

Let's assume the object distance \(u_{1}\) is the first point where the lens can be placed. Now set up the lens equation (\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]) for this point. Considering the direction towards the screen as positive, the total distance (\(70 cm\)) will be split into \(v_{1}\) and \(u_{1}\). Thus we can use the equation as \[ \frac{1}{f} = \frac{1}{70 - u_{1}} - \frac{1}{u_{1}} \]
03

Solve the Equation for \(u_{1}\)

Plug in the given values of \(f\) in the equation and solve for \(u_{1}\) to find the first point where the lens can be placed. After simplification, we obtain \(u_{1}\) to be approximately 44.9 cm and \(v_{1}\) to be approximately 25.1 cm.
04

Set Up the Lens Equation for the Second Point

Similarly, for the second point, the total distance will be split into \(70 - u_{2} = v_{2}\) and \(u_{2}\). So, \[ \frac{1}{f} = \frac{1}{70 - u_{2}} - \frac{1}{u_{2}} \]. Solve this equation by substituting given values.
05

Solve the Equation for \(u_{2}\)

After simplification, we obtain the second object distance as approximately 25.1 cm and the image distance as approximately 44.9 cm. This is the case of the lens forming a virtual image because now the lens is nearer to the screen and the perceived candlelight on the screen seems to originate from a point further than the screen.

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Most popular questions from this chapter

Show that identical objects placed equal distances on either side of the focal point of a concave mirror or converging lens produce images of equal size. Are the images of the same type?

A converging lens has focal length \(4.0 \mathrm{cm} .\) A 1.0 -cm-high arrow is located \(7.0 \mathrm{cm}\) from the lens with its lowest point \(5.0 \mathrm{mm}\) above the lens axis. Make a full-scale ray-tracing diagram to locate both ends of the image. Confirm using the lens equation.

Does a fish in a spherical bowl appear larger or smaller than it actually is?

If a lens suffers from spherical aberration, stopping down will a. worsen the focus. b. improve the focus. c. not affect the focus.

If you placed a screen at the location of a virtual image, would the image appear on the screen? Why or why not?
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