Question

A thick plane mirror shows a number of images out of the filament of an electric bulb. Of these, the brightest image is the
$ \left( A \right){1^{st}} \\
  \left( B \right){2^{nd}} \\
  \left( C \right){3^{rd}} \\
  \left( D \right){4^{th}} \\ $

Answer 1

Hint :To solve this question, first construct a figure showing the reflection of light from a thick mirror. Then consider all the images that will be formed and then the brightness of each is compared using the concepts of reflection.
Consider the four images formed. Compare the origins of their formation and then the brightness.

Complete Step By Step Answer:

In the mirror $ M $ , as we can see that the reflection takes place at the two surfaces $ AB $ and $ CD $ , the images are formed due to the reflection from both. Now as we know that the second surface is the silvered one, then the image formed due to the second one i.e. $ CD $ is more bright as compared to that from the first surface $ AB $ . The further images are formed due to the multiple reflections which involve the absorption of light from the medium. This results in the brightness of the images getting decreased, therefore, we can say that the further images become fainter due to the light absorption.
So, following this concept, we infer here that the brightest image formed will be the second image which is formed by the silvered surface.
Hence, option $ \left( B \right) $ is the correct one.

Note :
The images first, third and fourth are fainter, where third and fourth are due to absorption where first is less bright as the surface is not silvered. However, if the surface $ AB $ were silvered, then only one image would have been formed.