Fig shows an object and its image formed by a thin lens. Then the nature and focal length of lens is

f = 4.8

cm converging lens

No worries! We‘ve got your back. Try BYJU‘S free classes today!

f = - 4.8

cm diverging lens

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

f = 2.18

cm converging lens

No worries! We‘ve got your back. Try BYJU‘S free classes today!

f = - 2.18

cm diverging lens

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $f = - 4.8$ cm diverging lens
From figure, $u = - (3 + 5) = - 8$ cm, $v = - 3$ cm

Using lens formula, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$∴$ $\frac{1}{- 3} - \frac{1}{- 8} = \frac{1}{f}$

$⟹ f = - 4.8$ cm

As the focal length of the lens comes out to be negative, thus its a diverging lens (concave lens).

Suggest Corrections

Similar questions

Q. An object is kept

60

c m

from a lens gives a virtual image

20

c m

in front of lens. What is the focal length of the lens? Is it converging lens or diverging lens?

A diverging lens, focal length $f_{1} = 20 cm$ is separated by 5 cm from a converging mirror, focal length $f_{2} = 10 cm$ . Where should an object be placed from the lens so that a real image is formed at the object itself ?

Q. An object is placed 20 cm from (a) a converging lens, and (b) a diverging lens, of focal length 15 cm. Calculate the image position and magnification in each case.

Q. A diverging lens with magnitude of focal length

25 c m

is placed at a distance of

15 c m

from a converging lens of magnitude of focal length

20 c m

. A beam of parallel light falls on the diverging lens. The final image formed is :

Q. A diverging lens with magnitude of focal length

25 c m

is placed at a distance of

15 c m

from a converging lens of magnitude of focal length

20 c m

. A beam of parallel light falls on the diverging lens. The final image formed is

Join BYJU'S Learning Program

Fig shows an object and its image formed by a thin lens. Then the nature and focal length of lens is

The correct option is B f=−4.8 cm diverging lensFrom figure, u=−(3+5)=−8 cm, v=−3 cmUsing lens formula, 1v−1u=1f∴ 1−3−1−8=1f ⟹f=−4.8 cmAs the focal length of the lens comes out to be negative, thus its a diverging lens (concave lens).