Question

Two students in a physics laboratory each have a concave mirror with the same radius of curvature, 50 cm. Each student places an object in front of a mirror. The image in both mirrors is three times the size of the object. However, when the students compare notes, they find that the object distances are not the same. What is the distance of the closer object?

Answer 1

16.67 cm.

Step-by-step explanation:

radius of curvature r = 50 cm

Focal length f = 25 cm

Magnification can be achieved 3 times in concave mirror in two ways .

1) By putting the object very close to the mirror ( within focal length ) so that magnified virtual and erect image is observed.

2) By putting the object between f and 2f so that real and inverted magnified image is observed .

In the question , the object distance in the former case has been asked.

\frac{v}{u}=3\\

v = 3u

Using the mirror formula we can calculate the required distance as follows

\frac{1}{v}+\frac{1}{u}=\frac{1}{f}

Put v = -3u, f = -25.

-\frac{1}{3u}+\frac{1}{u}=\frac{1}{-25}\\

u=16.67cm

Answer 2

The distance of the closer object 16.67 cm.

Magnification of the mirror

The magnification of the mirror is calculated as follows;

M = v/u

3 = v/u

v = 3u

Object distance

The distance of the object in the converging mirror is calculated as follows;

1/f = 1/u + 1/v

where;

• f is the focal length = half of radius of curvature = 0.5 x 50 cm = 25 cm

1/25 = 1/u + 1/3u

1/25 = 4/3u

3u = 25 x 4

3u = 100

u = 100/3

u = 33.33 cm

Distance of closer object = 50 cm - 33.33 cm = 16.67 cm

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