Question:

A converging lens with a focal length of 15 cm is placed in front of a converging mirror with a radius of curvature of 30 cm. The object is placed 20 cm in front of the lens.

a) Determine the position and magnification of the image formed by the lens-mirror system. b) Draw a ray diagram to illustrate the formation of the image. c) Calculate the height of the image if the height of the object is 4 cm.

Answer:

a) To find the position and magnification of the image formed by the lens-mirror system, we need to consider the individual contributions of the lens and the mirror.

First, let's find the position and magnification due to the lens: Using the lens equation: 1/f_lens = 1/do + 1/di

Where f_lens is the focal length of the lens, do is the object distance, and di is the image distance.

Given: f_lens = 15 cm do = -20 cm (negative because the object is placed in front of the lens)

Substituting the values: 1/15 = 1/-20 + 1/di

Simplifying the equation: 1/di = 1/15 + 1/20 1/di = (4 + 3) / 60 1/di = 7/60

Taking the reciprocal: di = 60/7 cm

Next, let's find the position and magnification due to the mirror:

Using the mirror formula: 1/f_mirror = 1/do + 1/di

Where do is the object distance from the mirror, and di is the image distance from the mirror.

Since the object is already placed in front of the lens, the distance from the mirror to the object is the same as the distance from the lens to the object, which is -20 cm.

Given: f_mirror = R_mirror/2 = 30/2 = 15 cm do_mirror = -20 cm (same as object distance for lens)

Substituting the values: 1/15 = 1/-20 + 1/di_mirror

Simplifying the equation: 1/di_mirror = 1/15 + 1/20 1/di_mirror = (4 + 3) / 60 1/di_mirror = 7/60

Taking the reciprocal: di_mirror = 60/7 cm

Since the object is placed between the lens and mirror, we need to switch the signs of the distances for the mirror:

di_mirror = -60/7 cm

To find the final position and magnification, we need to consider the combined lens-mirror system:

Using the lens formula for the combined system: 1/f_total = 1/do + 1/di_total

Given: do = -20 cm (same as object distance for lens) di_total is the final image distance, which is the distance from the lens to the image formed by the lens-mirror system.

Substituting the values: 1/f_total = 1/-20 + 1/di_total 1/f_total = -1/20 + 1/(60/7)

Simplifying the equation: 1/f_total = -1/20 + 7/60 1/f_total = (-3 + 7) / 60 1/f_total = 4/60 1/f_total = 1/15

Taking the reciprocal: f_total = 15 cm

Therefore, the final image formed by the lens-mirror system is located at a distance of 15 cm from the lens.

Now, let's calculate the magnification of the image:

Using the magnification formula: magnification = -di/do = -di_total/do

Given: do = -20 cm (same as object distance for lens) di_total = 15 cm (distance from the lens to the final image)

Substituting the values: magnification = -(15) / (-20) magnification = 3/4

Therefore, the magnification of the image formed by the lens-mirror system is 3/4.

b) The ray diagram for the formation of the image would show two rays: one passing through the lens and one reflecting off the mirror. The diagram would depict the rays starting from the object, passing through the lens, and then reflecting off the mirror to form the image.

c) To calculate the height of the image, we can use the magnification formula:

magnification = -di/ do = hi/ho

Given: ho (height of the object) = 4 cm magnification = 3/4

Substituting the values: 3/4 = -di/ (-20) 3/4 = -di / -20

Cross-multiplying: 4*(-di) = 3*(-20) -4*di = -60 di = 15 cm

Now, to find the height of the image (hi), we use the magnification formula:

hi = magnification * ho = (3/4) * 4 hi = 3 cm

Therefore, the height of the image formed by the lens-mirror system is 3 cm.