Question

the image produced by a concave mirror is one quarter the size of the object. if the object is moved 5cm closer to the mirror​ the image will only half the size of the object. Find the focal length of the mirror

Answer 1

Answer:

Explanation:

Suppose u = initial distance of object

magnification, m = 1/4

=> Now, Magnification of a lens: m = v/u

1/4 = v/u

v = u/4

=> By putting the value of v in mirror formula, we get

1/v + 1/u = 1/f

(-4 / u) + (-1/u) = 1/f ...(1)

=> Now, if the object is moved 5cm closer to the mirror​ the image will only half the size of the object.

Thus, u' = -(u - 5)

m' = 1/2

Thus, m' = v'/u'

1/2 =v' / -(u - 5)

v' = -(u - 5) / 2

=> By putting the value of v' in mirror formula, we get

1/v' + 1/u' = 1/f

(-2 / u-5) + (-1/ u-5) = 1/f ...(2)

From eq (1) and (2) ,

(4 / u) + (1/u) = (2 / u-5) + (1/ u-5)

5/u = 3/ u-5

5(u-5) = 3u

5u - 25 = 3u

2u = 25

u = 12.5

=> By substuting the value of u in eq (1), we get

(-4 / 12.5) + (-1/12.5) = 1/f

f = -2.5

here -ve sign indicates concave mirror

Thus,  the focal length of the mirror is -2.5

Answer 2

Explanation:

ANSWER

In first case, if object distance is x, image distance =x/4.

While it is 2

nd

case, object distance becomes (x−5 cm) and image distance (x−5 cm)/2. Using mirror formula we get,

In first case, =

f

1

=−

x

5

In second case,

f

1

=−

(x−5)

3

Solving we get, x=12.5 cm and f=−2.5 cm and hence ∣f∣=2.5 cm