Question

A diverging lens with \(f=-30.0 \mathrm{~cm}\) is placed \(15.0 \mathrm{~cm}\) behind a converging lens with \(f=20.0 \mathrm{~cm}\). Where will an object at infinity in front of the converging lens be focused?

Short answer

When an object is at infinity in front of a converging lens with a focal length of 20.0 cm and a diverging lens with a focal length of -30.0 cm placed 15.0 cm apart, the final image will be focused at 8.0 cm in front of the converging lens.

Step-by-step solution

Find the image formed by the converging lens

Since the object is at infinity, the rays incident on the converging lens will be parallel to the principal axis. Therefore, the image will be formed at the focal point of the converging lens. Thus, the image formed by the converging lens has a distance of \(20.0 \mathrm{~cm}\).

Calculate the object distance for the diverging lens

As the converging lens and diverging lens are \(15.0 \mathrm{~cm}\) apart, the object distance for the diverging lens (\(s_2\)) is the difference between the image distance from the converging lens (\(s_1\)) and the distance between the lenses: $$ s_2 = s_1 - 15.0 \mathrm{~cm} = 20.0 \mathrm{~cm} - 15.0 \mathrm{~cm} = 5.0 \mathrm{~cm} $$

Apply the lens formula to the diverging lens

We have the focal length of the diverging lens (\(f_2=-30.0 \mathrm{~cm}\)) and the object distance (\(s_2 = 5.0 \mathrm{~cm}\)). We can use the lens formula to find the image distance from the diverging lens (\(s_2'\)): $$ \frac{1}{f_2} = \frac{1}{s_2} + \frac{1}{s_2'} $$ Plug in the values for \(f_2\) and \(s_2\): $$ \frac{1}{-30.0 \mathrm{~cm}} = \frac{1}{5.0 \mathrm{~cm}} + \frac{1}{s_2'} $$

Solve for the image distance (\(s_2'\))

Next, solving for the image distance \(s_2'\): $$ \frac{1}{s_2'} = \frac{1}{-30.0 \mathrm{~cm}} - \frac{1}{5.0 \mathrm{~cm}} = -\frac{1}{30.0 \mathrm{~cm}} - \frac{1}{5.0 \mathrm{~cm}} $$ $$ s_2' = \frac{-1}{\frac{-1}{30.0 \mathrm{~cm}} -\frac{1}{5.0 \mathrm{~cm}}} = -12.0 \mathrm{~cm} $$

Calculate the final image position

Now we have the image distance from the diverging lens as \(s_2' = -12.0 \mathrm{~cm}\). This is a virtual image on the same side of the object (hence the negative sign). To find the final image position from the converging lens, we need to add the distance between the lenses to \(s_2'\): $$ s_{\mathrm{final}} = s_1 + s_2' = 20.0 \mathrm{~cm} + (-12.0 \mathrm{~cm}) = 8.0 \mathrm{~cm} $$ So, the object at infinity in front of the converging lens will be focused at \(8.0 \mathrm{~cm}\) in front of the converging lens.