An object that is 5.0 cm tall is placed 15.0 cm to the left of a converging lens

Question

An object that is 5.0 cm tall is placed 15.0 cm to the left of a converging lens that has a focal length of 10.0 cm. A plane mirror is located 40.0 cm to the right of the lens. The scale for the figure is 1 box = 5cm. First a ray diagram is drawn to find the location of the image (image 1) created by the lens, while ignoring the mirror. After passing through the lens, the light then encounters the mirror and a second image (image 2) is formed. In this case, image 1 is used as the object and a ray diagram is drawn again to find the location of image 2. Lastly, the mirror sends the light back towards the lens and the lens creates another image (image 3). We use image 2 as the object in this situation. A ray diagram should be drawn to find the location of image 3.

a. Find the location of image 3 using ray diagrams

b. Mark a point on the ray diagram where your eye could be so that you could see image 3.

Explanation / Answer

for lens forming image 1,

1/f = 1/u + 1/v

1/v = 1/10 - 1/(-15)

v = 30/5 = 6 cm

Magnification of image = -v/u = h2/h1

h2 = (-6)/(-15) * 5 = 2 cm with arrow in the downward direction of the principal axis.

Image 2 formation,

for the case of plane mirror, object distance = image distance and

object hieght = image hieght.

so, u = v = 40 - 6 = 34 cm in the right side of the plane mirror of image hieght = 2 cm.

or Image 3 formation,

u = 40 + 34 = 74 cm

1/f = 1/u + 1/v

1/v = 1/(-10)- 1/74

v = -740/84 = -8.81 cm

M = -v/u = h2/h1

h2 = 8.81/74 * 2 = 0.24 cm

So, image 3 will form on the left side of the lens at a distance of 8.81 cm of size 0.24 cm with arrow in the upward direction above the principal axis.

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