0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

A thin equiconvex lens is made of a glass of refractive index 1.5 and its focal length is 0.2 m. If it acts as a concave lens of 0.5 m focal length when dipped in a liquid, the refractive index of the liquid is:
  1. 178
  2. 158
  3. 138
  4. 98

A
98
B
138
C
178
D
158
Solution
Verified by Toppr

Given,
μ1=1.5f1=0.2mf2=(0.5)μ2=?

Using lens makers formula and taking ratios, f1f2=(μ21)(μ11)

0.2(0.5)=(1.5μe1)(1.51)

0.2×0.50.5=(11.5μe)

1.5μe=0.8

μe=158

Was this answer helpful?
11
Similar Questions
Q1
A thin equiconvex lens is made of a glass of refractive index 1.5 and its focal length is 0.2 m. If it acts as a concave lens of 0.5 m focal length when dipped in a liquid, the refractive index of the liquid is:
View Solution
Q2
A thin bi-convex lens of focal length 20 cm is made of glass of refractive index 1.5. When it is dipped in a liquid of refractive index 98, it acts as a:
View Solution
Q3
A thin equiconvex lens is made of glass of R.I 1.5 and its focal length is 0.2 If it acts a concave lens of 0.5 m focal length when dipped in a liquid, the R.I of liquid is
View Solution
Q4
A thin converging lens is made up of a glass of refractive index 1.5. It acts like a concave lens of focal length 50 cm when immersed in a liquid of refractive index (158) The focal length of the converging lens in air is (in m)
View Solution
Q5
A thin coverging lens is made up of glass of refractive index 1.5. It acts like a concave lens of focal length 50 cm, when immersed in a liquid of refractive index 158, focal length of converging lens in air (in metre) is
View Solution