Question

A lens has a power of $10D$, when placed in air. When it is immersed in water $\left( {u = \dfrac{4}{3}} \right)$ . The change in power is (refractive index of lens material is $1 \cdot 5$.
A.$2 \cdot 55D$
B. $5 \cdot 0D$
C. $7 \cdot 5D$
D. $6 \cdot 67D$

Answer 1

Hint: We have to find the change in power. For solving this question we have to know the focal length which is $p = \dfrac{1}{f}$ so when a lens is immersed in water then the focal length increased and we know that focal length is inversely proportional to the angular magnification so that the lens immersed in water that decreases the power. The power of lens is the reciprocal of its focal length. And then change the power of refractive index of lens material. Also, we have to find when it is placed in air. So, substituting all these values we will get the required answer.

Complete step by step answer:
We know that,
$D = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Where, D is Dioptre
$ = \left( {\mu - 1} \right) \cdot K$
Now, put the value,
The refractive index of a material is $1 \cdot 5$ given.
$10 = \left( {1 \cdot 5 - 1} \right) \times K$
Now, putting the value of $K = \dfrac{{10}}{{0 \cdot 5}}$
$ = 20$
Now, put the value of D in the following equation after it immersed in water.
The refractive index of is $\dfrac{4}{3}$ .
Now, we have to calculate the power,
Now, putting all the value.

$D = \left( {\dfrac{4}{3} - 1} \right) \times K$
$ = \dfrac{1}{3} \times 20$
$ = \dfrac{{20}}{3}$
Substituting all the values we get.
$ = 6 \cdot 67D$
Hence, the correct option is D. Which is required answer.

Note: Since, we know that when a lens is dipped in water its automatically focal length increases. Lens is a thin curved piece of glass. For solving focal length is also defined as it is the distance between the lens and the image sensor when the subject is usually used. Refractive index can be defined as the distance from a focal point of the lens or mirror to the corresponding plane.
The refractive of the k=lens can be defined as the higher the refractive index the slower the light travels, Put the value of $\mu $and D in the given equation. So, these are some important steps which we have to remember to get the required answer.