Question

A concave mirror of radius $40cm$ lies on a horizontal table and water is filled in it up to a height of $5.00cm$. A small dust particle floats on the water surface at a point $P$ vertically above the point of contact of the mirror with the table. The image of the dust particle as seen from a point directly above it is $8+\frac{a}{4}cm$. Find the value of $a$. [Use the refractive index of water is $1.33$].

Answer 1

Given: A concave mirror of radius 40 cm lies on a horizontal table and water is filled in it up to a height of 5.00 cm. A small dust particle floats on the water surface at a point P vertically above the point of contact of the mirror with the table. The image of the dust particle as seen from a point directly above it is $8+\frac{a}{4}cm$

To find the value of $a$

Solution:

The ray diagram is shown in above fig.

According to the given criteria,

Radius of curvature of the mirror, $R=40$

Therefore, focal length of the mirror, $f=\frac{R}{2}=\frac{40}{2}=20cm$

The object distance, $u=5cm$

Applying mirror formula,

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}⟹\frac{1}{20}=\frac{1}{v}+\frac{1}{5}⟹\frac{1}{v}=\frac{1}{20}-\frac{1}{5}⟹\frac{1}{v}=\frac{1-4}{20}⟹v=-6.67cm$

The negative sign shows that the image is formed on the other side of the mirror, i.e., below the mirror and hence is virtual.

These reflected rays are refracted at the water surface and go to the observer. The depth of the point $A_1$ from the surface is $6.7+5=11.7cm$.

Due to refraction at the water surface, the image $A_1$ will be shifted above by a distance,

$\left(11.7\right)\left(1-\frac{1}{1.33}\right)=2.925cm$

Thus the final image is formed at a point $(11.7-2.925)=8.775cm$ below the water surface.

But given theimage of the dust particle as seen from a point directly above it is $8+\frac{a}{4}cm$

equating this with the obtained value, we get

$8+\frac{a}{4}cm=8.775⟹\frac{a}{4}=0.775⟹a=3.1$