A double convex thin lens made of glass of refractive index $1.6$ has radii of curvature $15 c m$ each. The focal length of this lens when immersed in a liquid of refractive index $163$ is:

- 407 c m

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

250 c m

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

125 c m

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

25 c m

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

Open in App

Solution

The correct option is C $- 407 c m$
As we know,
$\frac{f_{l}}{f_{a}} = \frac{(_{a} μ - 1)}{(_{l} μ - 1)}; f_{a} = \frac{R}{2 (μ_{g} - 1)} = \frac{15}{2 (1.6 - 1)} = 12.5$
$\frac{f_{1}}{12.5} = \frac{(1.6 - 1)}{(\frac{1.6}{1.63} - 1)} \Rightarrow f + 1 = - 407 c m$

Suggest Corrections

Similar questions

A double convex thin lens made of glass of refractive index 1.6 has radii of curvature 15 cm each. The focal length of this lens when immersed in a fluid of refractive index 1.63 is

Q. A thin double convex lens has radii of curvature each of magnitude 40 cm and is made of glass with refractive index 1.65. Its focal length is nearly

Q. A thin convex lens of refractive index

1.5

has

20 c m

focal length in air. If the lens is completely immersed in a liquid refractive index

1.6

, its focal length will be :

Q. A thin convex lens which is made of glass (refractive index

1.5

) has a focal length of

20 c m

.it is now completely immersed in a transparent liquid having refractive index

1.75

. Find the new focal length of the lens.

Join BYJU'S Learning Program

A double convex thin lens made of glass of refractive index 1.6 has radii of curvature 15cm each. The focal length of this lens when immersed in a liquid of refractive index 163 is:

The correct option is C −407cmAs we know,flfa=(aμ−1)(lμ−1); fa=R2(μg−1)=152(1.6−1)=12.5f112.5=(1.6−1)(1.61.63−1)⇒f+1=−407 cm

A double convex thin lens made of glass of refractive index $1.6$ has radii of curvature $15 c m$ each. The focal length of this lens when immersed in a liquid of refractive index $163$ is:

The correct option is C $- 407 c m$
As we know,
$\frac{f_{l}}{f_{a}} = \frac{(_{a} μ - 1)}{(_{l} μ - 1)}; f_{a} = \frac{R}{2 (μ_{g} - 1)} = \frac{15}{2 (1.6 - 1)} = 12.5$
$\frac{f_{1}}{12.5} = \frac{(1.6 - 1)}{(\frac{1.6}{1.63} - 1)} \Rightarrow f + 1 = - 407 c m$