Question

A point luminous object (O) is at a distance h from the front face of a glass slab of width d and of refractive index $\mu $. On the back face of a slab is a reflecting plane mirror. An observer sees the image of an object in the mirror as shown in the figure. Distance of image from front face as seen by observer will be:

A. $h + \dfrac{{2d}}{\mu }$
B. $2h + 2d$
C. $h + d$
D. $h + \dfrac{d}{\mu }$

Answer 1

Hint:In the given question, we have to find the apparent depth of the luminous object. In the question it is given that on a mirror there is a glass slab of width d and from the front face of the glass slab there is a point object at a distance of h. And the observer is also seeing from the same side as the object. Here, the observer will see the mirror uplifted from its original position since the rays coming from the mirror will get refracted upon passing through the glass slab and making the observer perceive that the mirror is uplifted.

This is called apparent depth. The observer will perceive the mirror at an apparent depth and this apparent depth of the mirror will be given by $\dfrac{d}{\mu }$ . Now, we can find the distance of the object from the virtual image of the mirror and we know that the image of object O will be at the same distance behind the apparent mirror as in front, thus we can find the distance of the image from the front face of the slab.

Complete step by step answer:
The mirror is placed under a glass slab, so it will appear to be uplifted from its original position, so the apparent depth of the mirror will be given by:
\[{\text{Apparent depth = }}\dfrac{{{\text{Real depth}}}}{{{\text{refractive index of the medium}}}}\]
Hence,
\[ \Rightarrow {\text{Apparent depth = }}\dfrac{d}{\mu }\]
So, we can say that the virtual image of the mirror MM will be formed at distance $\dfrac{d}{\mu }$ from the front face of the glass slab.

Now, as we know the virtual image of an object in front of the plane mirror is at the same distance as it is in front.
So, we can see here that the point object O is at a distance $\left( {h + \dfrac{d}{\mu }} \right)$ from the virtual mirror mm’.
So, its image $I$ will also be at a distance of $\left( {h + \dfrac{d}{\mu }} \right)$ behind the mirror mm’.
The distance of image from the front face of the glass slab would be given by:
$
x = \left( {h + \dfrac{d}{\mu }} \right) + \dfrac{d}{\mu } \\
\therefore x = h + \dfrac{{2d}}{\mu } \\
$
Hence, the option A is correct.

Note:The apparent depth is the distance of the virtual image from the front face while the real depth is the distance of the real object from the front face of the slab. Some students may approach this question as since the object is at a distance of $\left( {h + d} \right)$ its image will be formed $\left( {h + d} \right)$ distance behind the mirror and this virtual image of the object will get shifted by $d\left( {1 - \dfrac{1}{\mu }} \right)$ when its rays reach up to observer after refraction.

But this will get us to the wrong answer $\left( {h + \dfrac{d}{\mu }} \right)$ . The mistake here is that we have neglected the fact that the rays will get refracted two times: first when rays from the object reaches to the mirror and second when the rays from the object’s virtual image reach up to the observer. So, there is two times shift of $\dfrac{d}{\mu }$ , giving us correct answer $\left( {h + \dfrac{{2d}}{\mu }} \right)$ .