Question

.
At what distance from a concave mirror of focal
length 10 cm must an object be placed in order
that an image double its size may be obtained ?​

Answer 1

Answer:

Let the object size be X

then image size becomes 2X

by applying formula we get:-

1/f=1/v+1/u

1/-10=1/2x+1/x

1/-10=3/2x

X= -15 cm

now the image size comes..... -15 ×2= -30

Answer 2

f = - 10 cm (w/ sign convention)

m = ± 2 (as it is not specified about the type of image so formed)

Let, object distance = u

Now, we know:

- v/u = m & mirror formula : 1/f = 1/v + 1/u

Case 1 (+ve mag):

=> - v/u = + 2

=> v = - 2u

∴ 1/(- 10) = 1/(- 2u) + 1/u

=> 1/(- 10) = - 1/2u + 1/u

=> u = (1 - 1/2)(- 10)

=> u = (1/2)(- 10) = - 5.00 cm

(for virtual and erect image)

Case 2 (-ve mag):

=> - v/u = - 2

=> v = 2u

∴ 1/(- 10) = 1/2u + 1/u

=> u = (1/2 + 1)(- 10)

=> u = (3/2)(- 10)

=> u = - 15.00 cm

(for real and inverted image)

Therefore, for a twofold sized image, the object should be kept either 5 cm or 15 cm away (before) the mirror.