Question

A composite slab consisting of different media is placed in front of a concave mirror of radius of curvature 150 cm as shown in Fig. The whole arrangement is immersed in water. An object O is placed at a distance 20cm from the slab. The R.I. of different media are given in the diagram. Find the final position of the image formed by the system

Answer 1

The bending of a ray of light as it passes through one medium and through another is measured by the refractive index, also known as the index of refraction. If I is the angle of incidence of a ray in vacuum (the angle between the incoming ray and the natural, which is perpendicular to the surface of a medium), and r is the angle of refraction (angle between the ray in the medium and the normal), The refractive index n is equal to the ratio of the sine of the angle of incidence to the sine of the angle of refraction; it is also equal to the velocity of light c in empty space divided by its velocity v in a material.

Complete step-by-step solution:
In front of a concave mirror with a radius of curvature of 150 cm, a composite slab made of various media is mounted. The whole setup is submerged in water. We need to find the final position of the image formed
We know that in Mirror or lens formula when we write
$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$
Here, “u” “v”, “f” are distances of object image and focal length respectively, but this is the case where we are doing the whole procedure in a medium of refractive index 1, and that is why we have written (1) in the numerator, which stands for the refractive index
The actual formula will look like
$\dfrac{R.I.}{u}+\dfrac{R.I.}{v}=\dfrac{R.I.}{f}$
Here, R.I. stands for the refractive index
Now, since we have different refractive at each point and distance that is why we need to write the whole equation as
$x=(\dfrac{{{u}_{1}}}{R.I.}+\dfrac{{{u}_{2}}}{R.I.}+\dfrac{{{u}_{3}}}{R.I.}+\dfrac{{{u}_{4}}}{R.I.})\times R.I.+{{u}_{5}}$
Where “X” is the distance of object (4th image)
The image formed after 4 refraction would be taken as the object.
The distance of object (4th image) from mirror's pole is given by
$x=(\dfrac{20}{\dfrac{4}{3}}+\dfrac{54}{51}+\dfrac{24}{1.0}+\dfrac{45}{1.5})\times \dfrac{4}{3}+10$
$\Rightarrow x=\left( 15+1.06+24+30 \right)\times \dfrac{4}{3}+10=\left( 70.06\times \dfrac{4}{3} \right)+10$
$\Rightarrow x=103cm$
On solving this, we get
X = 103 cm
Hence, the final position of the image formed will be at 103 cm from the object which would be on the 54cm slab after the ray retracing its path.

Note: A mirror is a reflective surface that reflects light and creates either a real or a virtual image. When an object is positioned in front of a mirror, the mirror reflects the reflection of the same object. When a hollow sphere is split into pieces and the outer surface of each cut part is painted, it becomes a mirror, with the inner surface reflecting the light. A concave mirror is the name for this kind of mirror. As light hits and bounces back from the concave mirror reflecting surface, it converges at a point. As a result, it's often called a converging mirror.