Question

A thin, diverging lens with a focal length of 50.0 cm has the same principal axis as a concave mirror with a radius of 60.0 cm. The center of the mirror is 20.0 cm from the lens, with the lens in front of the mirror. An object is placed 27.0 cm in front of the lens.

Where is the final image due to the lens–mirror combination? Enter the image distance with respect to the mirror. Follow the sign convention.

A. 1800 cm
B. -1800 cm
C. 1600 cm
D. -1600 cm

Answer 1

The final image due to the lens-mirror combination is located in front of the mirror.

Step-by-step explanation:

To find the location of the final image due to the lens-mirror combination, we can use the mirror equation and the lens equation.

The mirror equation is given by 1/f_mirror = 1/d_mirror + 1/d_image_mirror, where f_mirror is the focal length of the mirror, d_mirror is the distance of the mirror from the object, and d_image_mirror is the distance of the image from the mirror.

The lens equation is given by 1/f_lens = 1/d_lens + 1/d_image_lens, where f_lens is the focal length of the lens, d_lens is the distance of the lens from the object, and d_image_lens is the distance of the image from the lens.

In this case, the lens and mirror have the same principal axis, so the image distance for the lens is the same as the object distance for the mirror. We can set d_image_mirror = d_lens and solve the two equations simultaneously to find the location of the final image.

Plugging in the given values, we have 1/60 = 1/20 + 1/d_image_mirror for the mirror equation and 1/50 = 1/27 + 1/d_image_lens for the lens equation. Solving these equations gives us d_image_mirror = -1600 cm and d_image_lens = 1800 cm.

Since the lens is in front of the mirror, the final image will be located in front of the mirror, so the answer is B. -1800 cm.