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Chapter 24: Problem 2

A converging lens and a diverging lens, separated by a distance of $30.0 \mathrm{cm},$ are used in combination. The converging lens has a focal length of \(15.0 \mathrm{cm} .\) The diverging lens is of unknown focal length. An object is placed \(20.0 \mathrm{cm}\) in front of the converging lens; the final image is virtual and is formed \(12.0 \mathrm{cm}\) before the diverging lens. What is the focal length of the diverging lens?

Short Answer

Expert verified
Answer: The focal length of the diverging lens is -60.0 cm.

Step by step solution

01

Understand the given information

We have a converging lens with a focal length of 15cm and a diverging lens with an unknown focal length. The object is placed 20 cm in front of the converging lens, and the final image is virtual, located 12 cm before the diverging lens. The lenses are separated by 30 cm.
02

Calculate the first image position using the lens formula for the converging lens

To find the position of the image formed by the converging lens, we can use the lens formula: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) Where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. For the converging lens: \(f=15.0\,cm\), \(d_o=20.0\,cm\) We can now solve for \(d_i\), \(\frac{1}{15.0} = \frac{1}{20.0} + \frac{1}{d_i}\)
03

Solve for the image distance \(d_i\) of the converging lens

We need to solve the equation for \(d_i\): Rearrange and solve for \(d_i\), \(\frac{1}{d_i} = \frac{1}{15.0} - \frac{1}{20.0}\) Find a common denominator and subtract the fractions: \(\frac{1}{d_i} = \frac{1}{60}\) Thus, \(d_i = 60.0\,cm\)
04

Calculate the object distance for the diverging lens

The distance from the converging lens image to the diverging lens will be the object distance for the diverging lens. Since lenses are 30 cm apart, and we have calculated the image distance \(d_i\) as 60 cm Object distance, \(d_o' = d_i - 30.0 = 60.0 - 30.0 = 30.0\,cm\)
05

Determine the image distance of the diverging lens

Given that the final image is virtual and is formed 12 cm before the diverging lens, it means the image distance for the diverging lens will be negative. Therefore, \(d_i'=-12.0\,cm\)
06

Apply the lens formula for the diverging lens to find the focal length

Again we apply the lens formula with the new object and image distances for the diverging lens to find its focal length, \(f'\): \(\frac{1}{f'} = \frac{1}{d_o'} + \frac{1}{d_i'}\) Plug in the values we found: \(\frac{1}{f'} = \frac{1}{30.0} + \frac{1}{-12.0}\)
07

Solve for the focal length of the diverging lens

Solve for \(f'\): \(\frac{1}{f'} = \frac{-1}{60}\) So, \(f' = -60.0\,cm\) The focal length of the diverging lens is \(-60.0\,cm\).

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Most popular questions from this chapter

A converging lens with a focal length of \(3.00 \mathrm{cm}\) is placed $24.00 \mathrm{cm}\( to the right of a concave mirror with a focal length of \)4.00 \mathrm{cm} .\( An object is placed between the mirror and the lens, \)6.00 \mathrm{cm}\( to the right of the mirror and \)18.00 \mathrm{cm}$ to the left of the lens. Name three places where you could find an image of this object. For each image tell whether it is inverted or upright and give the total magnification.
The eyepiece of a microscope has a focal length of \(1.25 \mathrm{cm}\) and the objective lens focal length is \(1.44 \mathrm{cm} .\) (a) If the tube length is \(18.0 \mathrm{cm},\) what is the angular magnification of the microscope? (b) What objective focal length would be required to double this magnification?
A converging lens with focal length \(3.00 \mathrm{cm}\) is placed $4.00 \mathrm{cm}$ to the right of an object. A diverging lens with focal length \(-5.00 \mathrm{cm}\) is placed \(17.0 \mathrm{cm}\) to the right of the converging lens. (a) At what location(s), if any, can you place a screen in order to display an image? (b) Repeat part (a) for the case where the lenses are separated by \(10.0 \mathrm{cm}.\)
A slide projector has a lens of focal length \(12 \mathrm{cm} .\) Each slide is \(24 \mathrm{mm}\) by \(36 \mathrm{mm}\) (see the figure with Problem 16). The projector is used in a room where the screen is \(5.0 \mathrm{m}\) from the projector. How large must the screen be?
An object is located \(16.0 \mathrm{cm}\) in front of a converging lens with focal length \(12.0 \mathrm{cm} .\) To the right of the converging lens, separated by a distance of \(20.0 \mathrm{cm},\) is a diverging lens of focal length \(-10.0 \mathrm{cm} .\) Find the location of the final image by ray tracing and verify using the lens equations.
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