Question

A point source of light B is placed at a distance L in front of the centre of a mirror of length 'd ' hung vertically on a wall.A man walks in front of the mirror along a line parallel to the mirror at a distance 2L from it as shown in the figure. The greatest distance over which he can see the image of the light source in the mirror is ??

A. d/2
B. d
C. 2d
D. 3d

Give the correct solution along with proper explanation. Don't Spam !! ​

Answer 1

Topic :-

Ray Optics

Given :-

A point source of light B is placed at a distance L in front of the centre of a mirror of length 'd' hung vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance 2L from it.

To Find :-

The greatest distance over which he can see the image of the light source in the mirror.

Methodology :-

Generally, in such type of questions we first draw the image of light source and then j०in it with ends of assumed greatest distance of observer and then apply some basic properties of the triangle to calculate the assumed greatest distance.

Solution :-

Name the mirror as PQ.

Let greatest distance over which he can see the image of the light source in the mirror be RS.

Now, draw the image of light source, B' at a distance of 'L' behind the mirror as it is a plane mirror and plane mirror forms images behind the mirror at the same distance as that of object.

Now, j०in B'R and B'S.

We can observe two triangles, ∆B'PQ and ∆B'SR forming here.

∠B' = ∠B' ( common angle )

∠Q = ∠R ( PQ and SR are parallel )

∠P = ∠S ( PQ and SR are parallel )

So, we can say that ∆B'PQ and ∆B'SR are similar triangles.

So,

( PQ / Height of ∆B'PQ ) = ( SR / Height of ∆B'SR )

PQ = d

Height of ∆B'PQ = L

SR = ?

Height of ∆B'SR = L + 2L = 3L

Substituting values,

( d / L ) = ( SR / (L + 2L) )

( d / L ) = ( SR / 3L )

SR = ( (d × 3L) / L )

SR = 3dL / L

SR = 3d

Answer :-

The greatest distance over which he can see the image of the light source in the mirror is 3d.

Hence, option D is correct option.

Answer 2

Consider a light beam from the point source at B meets the mirror at the point A at a distance $\sf{x}$ from the bottom end of the mirror. The reflected ray reaches the man when he is at the point D. The normal to the mirror at A passes through E and meets the path of the man at F. Here AF = 2L and AE = L.

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Consider the triangles ABE and ACE.

• ∠BAE = ∠CAE (angle of incidence = angle of reflection)

• ∠BEA = ∠CEA = 90°

• AE is a common side.

Thus the triangles ABE and ACE are congruent and so,

$\sf{\longrightarrow CE=BE=\dfrac{d}{2}-x}$

Since the triangles ACE and ADF are similar,

$\sf{\longrightarrow\dfrac{FD}{EC}=\dfrac{AF}{AE}}$

$\sf{\longrightarrow\dfrac{FD}{\frac{d}{2}-x}=\dfrac{2L}{L}}$

$\sf{\longrightarrow\dfrac{FD}{\frac{d}{2}-x}=2}$

$\sf{\longrightarrow FD=2\left(\dfrac{d}{2}-x\right)}$

$\sf{\longrightarrow FD=d-2x}$

So the range below B, over which the man can see the image will be,

$\sf{\longrightarrow GD=d-2x+\dfrac{d}{2}-x}$

$\sf{\longrightarrow GD=\dfrac{3d}{2}-3x}$

Well a range above B is also possible, and it is the same as this GD.

So the net range or the greatest distance will be,

$\sf{\longrightarrow \Delta=2GD}$

$\sf{\longrightarrow\Delta=2\left(\dfrac{3d}{2}-3x\right)}$

$\sf{\longrightarrow\Delta=3d-6x}$

The greatest distance is given when $\sf{x=0}$ which brings the point A at the bottom end of the mirror.

$\sf{\longrightarrow\Delta=3d-6(0)}$

$\sf{\longrightarrow\underline{\underline{\Delta=3d}}}$

Hence (D) is the answer.