Question

A concave lens forms the image of an object such that the distance between the object and the image is 10 cm .If the magnification of the image is 1/4 . The focal length of the lens will be?

Answer 1

Hey here is your answer ^_^

⭐ANSWER :

⭐It is given that the distance between object and image is 10 cm.

If you draw the figure, you'll realise that the distance between the object and image is actually
the difference of distances of the object from the lens and the distance of image from the lens.
(Since image and object is on same side)

That is, |u| - |v| = 10 cm|v| = |u| - 10Since the magnification is positive and less than 1, it's a virtual image. 

We know that magnification by a lens is given by,

m = v/u = (u-10)/u = 1/4

 4u - 40 = u

3u = 40

u = - 40/3 cm  (since it is in -x axis)

Therefore, v = -10/3 (since virtual image is formed in -x axis)

Hence, using lens formula,

1/f = 3/-10 - 3/-40

1/f = -12/40 + 3/40

1/f = -9/40

f = -40/9 cm  = -4.4 cm

CHEERZ!

Hope it helped you out ⭐^_^⭐

Thanks ⭐(^^)⭐

Answer 2

Object distance u=-10 cm
Magnification,m=1/4
therefore m=v/u=1/4
V/-10=1/4
4v=-10
V=-10/4 or -2.5 cm

Using lens formula

1/F=1/v-1/u
1/F=1/(-2.5)-1/(-10)
1/f=1/-2.5+1/10
1/f=-4/10+1/10 (by taking lcm2.5×4=10)
1/f=-3/10
F=-10/3
f=-3.33 cm (ANS)