Question

An object is placed 21cm in front of a concave mirror of radius 10cm. A glass slab of thickness 3cm and refractive index 1.5 is then placed close to the mirror in the space between the object and the mirror. The distance of the near surface of the slab from the mirror is 1cm. The final image from the mirror will be formed at:
A. 4.67cm
B. 6.67cm
C. 5.67cm
D. 7.67cm

Answer 1

Hint: The concepts of combination of lenses and mirror will be required to solve this problem. We must remember the sign conventions for concave mirror and the mirror formula of concave mirror, which is, $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$. The glass slab will cause a shift for any kind of ray passing through it. The shift due to the glass slab is given by ,$d=(1-\dfrac{1}{n})t$.

Step by step solution:
Let’s start by making a diagram of how the object’s rays move through the optical system of the glass slab and the concave mirror.

The object is at O and the rays are shown to be emitting from them. The ray upon interacting with the glass slab, changes direction and again upon leaving the glass slab, changes back it’s direction. We retrace back the emergent ray and mark the new position of the object as${{O}_{1}}$. Point P in the concave mirror refers to the center of the mirror, where the principal axis coincides with it. The thickness of the glass slab is given to be 3cm and the refractive index of the glass slab is given to be (n=1.5). The object’s initial distance is 21cm, hence the distance OP is 21cm.
Due to the glass slab, the object shifts closer to the mirror by a distance d. The general formula to find this distance d is, $d=(1-\dfrac{1}{n})t$. Therefore, $d=(1-\dfrac{1}{1.5})3cm\Rightarrow d=(\dfrac{0.5}{1.5})3cm\Rightarrow d=1cm$.

Therefore, the object shifts forward by 1cm due to the glass slab. Hence, the distance${{O}_{1}}P$ is 20cm.

Now, to find the image distance, we will use the mirror formula for concave mirror, which is, $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$. Here, the object distance will be the updated objected distance, that is u=20cm. To find the focal length, we will use the general formula using the radius of the mirror. $f=\dfrac{R}{2}\Rightarrow f=\dfrac{10cm}{2}\Rightarrow f=5cm$. However, both object distance and focal length will be negative, since they are on the left of the mirror and by convention the ones on the left are negative. Hence, putting in these values and conditions into the mirror formula, we get, $\dfrac{1}{v}+\dfrac{1}{-20}=\dfrac{1}{-5}\Rightarrow \dfrac{1}{v}=\dfrac{1}{-5}-\dfrac{1}{-20}\Rightarrow \dfrac{1}{v}=\dfrac{-3}{20}\Rightarrow v=\dfrac{-20}{3}cm$. Hence, the image distance is equal to -6.67cm or 6.67cm to the left of the mirror.

However, the ray will again pass through the glass slab. This will induce an additional distance of 1cm to the original path. Hence, the updated final image distance from the mirror is 6.67+1=7.67cm.

Note:
The initial ray from the object upon hitting the glass slab, makes an angle (i) with it. This is the incident angle that it makes with the glass slab will be the same corresponding angle with which it will leave the glass slab. Hence, upon retracing the ray emerging from the glass slab back towards the object, we see that the rays are parallel to each other. This is an important property of glass slab.