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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