Two long plane mirrors M_1 and M_2 are kept i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the angle of incidence on mirror $M_1$ be $i_1 = 30^o$. The angle of reflection is also $30^o$. The angle the reflected ray makes with the sur

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Let the angle of incidence on mirror $M_1$ be $i_1 = 30^o$. The angle of reflection is also $30^o$. The angle the reflected ray makes with the surface of $M_1$ is $90^o - 30^o = 60^o$.In the triangle formed by the two mirrors and the ray,the angle at the vertex is $40^o$. The angle at the base on $M_1$ is $60^o$. Thus,the angle at the base on $M_2$ is $180^o - (40^o + 60^o) = 80^o$.The angle of incidence on $M_2$ is $90^o - 80^o = 10^o$. Since $10^o The angle of reflection from $M_2$ is $10^o$. The angle this ray makes with the surface of $M_2$ is $90^o - 10^o = 80^o$.Now,consider the triangle formed by the ray and the mirrors. The angle at the vertex is $40^o$. The angle at the base on $M_2$ is $80^o$. The angle at the base on $M_1$ is $180^o - (40^o + 80^o) = 60^o$.Since the angle with the surface of $M_1$ is $60^o$,the angle of incidence is $90^o - 60^o = 30^o$. This is the same as the initial angle of incidence,meaning the ray will continue to reflect between the mirrors.However,the question asks for the number of reflections. Based on the geometry,the ray will strike $M_1$ again,then $M_2$,and so on. Given the mirrors are long,the ray will undergo a finite number of reflections until it exits the system. For an angle of $40^o$,the total number of reflections is calculated as $N = \frac{180^o}{	heta} = \frac{180^o}{40^o} = 4.5$. The ray will undergo $4$ reflections.

Two long plane mirrors $M_1$ and $M_2$ are kept inclined to each other. The angle between reflecting surfaces is $40^o$. $A$ light ray strikes mirror $M_1$ at an angle of $30^o$ with the normal. The maximum number of reflections the ray will undergo is:

Similar Questions

Two parallel mirrors $M_1$ and $M_2$ are separated by a distance $a$. An object is placed between them. What is the distance between the $n^{th}$ image formed by $M_1$ and the $n^{th}$ image formed by $M_2$?

In the image given below,the total number of reflections of the light ray before it exits the system will be

Light incident normally on a plane mirror attached to a galvanometer coil retraces its path as shown in the figure. $A$ current in the coil produces a deflection of $3.5^{\circ}$ of the mirror. What is the displacement (in $cm$) of the reflected spot of light on a screen placed $1.5\; m$ away?

By what angle should $M_2$ be rotated,so that the light ray after reflection from both mirrors becomes horizontal? (The angle of incidence at mirror $M_1$ is $40^{\circ}$ and the angle between $M_2$ and the horizontal is $25^{\circ}$ as shown in the figure.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module