Question

A convex lens made of glass $\left( {{\mu _g} = 3/2} \right)$ has focal length $f$ in air. The image of an object placed in front of it is inverted, real and magnified. Now the whole arrangement is immersed in water $\left( {{\mu _w} = 4/3} \right)$ without changing the distance between the object and lens. Then
(A) The new focal length will become $4f$
(B) The new focal length will become $f/4$
(C) New image will be virtual and magnified
(D) New image will be real, inverted and smaller in size

Answer 1

Hint:Here first we have to find the focal length in water and air. Then we can say what type of an image is formed.

Complete step by step answer:
The focal length is positive for a convex lens and is the distance at which a ray of collimated light is centred on a single spot. The relationship between focal length and refractive index is given by the lens maker’s formula. The formula of the lens manufacturer is the relationship between the focal length of a lens to the material’s refractive index and the curvature of radii of its two surfaces. It is used by manufacturers of lenses to produce special power lenses from the glass of a specified refractive index.
Mathematically the lens maker’s formula is given by-
$\dfrac{1}{f} = \left( {\mu - 1} \right) \times \left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Where, $f$ is the focal length, $\mu $ is the refractive index and ${R_1}$ and ${R_2}$ are the radius of curvature of both surfaces. Either real or virtual images can be formed by convex lenses. For the image to be magnified in a convex lens the object should always be placed between the two focal lengths.
Given, a convex lens made of glass $\left( {{\mu _g} = 3/2} \right)$ has focal length $f$ in the air. The image of an object placed in front of it is inverted, real and magnified. Now the whole arrangement is immersed in water $\left( {{\mu _w} = 4/3} \right)$ without changing the distance between the object and lens.
Let the focal length be $f$ in air.
According to lens maker’s formula-
$\dfrac{1}{f} = \left( {\dfrac{3}{2} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Now, let the focal length be $f'$ in water.
$\dfrac{1}{{f'}} = \left( {\dfrac{{3/2}}{{4/3}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}
{{{R_2}}}} \right)$
Hence, from the above equations we get
$f' = 4f$
Since, the actual image is real and magnified, the object is located between $f$ and $2f$.
Now the focal length has changed to $4f$.
Hence,
$u = - u(f < u < 2f) \\
= u < f' \\ $
Therefore, the final image is virtual and magnified.
So, option C will be the correct answer.

Note:While using the lens maker’s formula we have to make sure that the refractive indices are put in the correct places. If we revert the refractive indices for air and water, the answer will be completely different. So, we have to be careful while doing that.