Question
Question: The focal length of a lens of refractive index \(\dfrac{3}{2}\) is 10cm in the air. The focal length...
The focal length of a lens of refractive index 23 is 10cm in the air. The focal length of that lens in a medium of refractive index 57 is:
a) -70 cm
b) 710 cm
c) 70 cm
d) None of these
Solution
Let us assume that the focal length of a lens in air is f1 and the focal length of the lens in a medium is f2. Now apply the maker's formula for each focal length in terms of radius, then divide both the equations to cancel out the terms and get the value of f2.
Formula used:
The lens maker’s formula for thin lens is used: f1=(μ−1)(R11−R21)
Here, R1 and R2 are the radius of curvature of the lens.
f is the focal length of the lens.
μ is the refractive index of the lens.
Complete step by step answer:
As we have the following data for air as a medium:
⇒f1=10⇒μ1=23
So, by applying maker’s formula, we get:
⇒101=(23−1)(R11−R21)......(1)
Also, we have the refractive index of lens in another medium as:
⇒μ2=57
So, by applying maker’s formula, we get:
⇒f21=(57−1)(R11−R21)......(2)
Now, divide equation (1) by equation (2), we get:
⇒f21101=(57−1)(R11−R21)(23−1)(R11−R21)⇒101f2=21×25⇒f2=45×101⇒f2=12.5cm
Hence, option (d) is the correct answer.
Additional Information:
The limitations for using the lens maker’s formula are:
-The lens must be thin because the separation between the two refracting surfaces will also be small.
-The medium on both sides of the lens needs to be the same.
Note: In this type of questions,major mistakes occurs while taking the sign convention so you should always remember that all the distances which are measured along the direction of incident rays are taken as positive and vice-versa.Also, you should remember that the distance measured upon the principal axis are taken as positive while the distances measured below the it is taken as negative.