Focal length of equiconvex lens with radius of each surface being $30 c m$ and refractive index $1.5$ , is

30 c m

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

60 c m

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

15 c m

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

7.5 c m

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

Open in App

Solution

The correct option is A $30 c m$
Using Lens maker's formula,
$\frac{1}{f} = (1.5 - 1) {\frac{1}{30} + \frac{1}{30}} = 0.5 \times \frac{2}{30} = \frac{1}{30}$
$f = 30 c m$

Suggest Corrections

Similar questions

R_{1} = 30 c m

R_{2} = 60 c m

μ = 1.5

μ = 1.5

1.6

60 c m

R_{1} = 30 c m

R_{2} = 60 c m ?

1.5

The radius of curvature for a convex lens is 40 cm, for each surface. Its refractive index is 1.5. The focal length will be

[MP PMT 1989]

Join BYJU'S Learning Program