Question

A lamp is placed \[6.0\,{\text{m}}\] from a wall. On putting a lens between the lamp and the wall at a distance of \[1.2\,{\text{m}}\] from the lamp, a real image of the lamp is formed on the wall. The magnification of the image is
A. $3$
B. $4$
C. $5$
D. $6$

Answer 1

Hint: Use the formula for the magnification of a lens. This formula gives the relation between the magnification of the lens, distance of the object from the lens and distance of the image from the lens. Calculate the distance of the lamp and distance of image of the lamp from the lens using the given information. Then calculate the magnification of the image of the lamp.

Formula used:
The magnification \[m\] of a lens is given by
\[m = - \dfrac{v}{u}\] …… (1)
Here, \[u\] is the distance of an object from the lens and \[v\] is the distance of the image of the object from the lens.

Complete step by step answer:
We have given that the distance between the lamp and the wall is \[6.0\,{\text{m}}\].
\[d = 6.0\,{\text{m}}\]
A lens is kept in between the lamp and the wall. The distance of the lamp from the lens is \[1.2\,{\text{m}}\]. This is the object distance.
\[u = 1.2\,{\text{m}}\]
We have asked to calculate the magnification of the image of the lamp on the wall.

Let us first calculate the distance of the image of the lamp from the lens.The distance \[d\] between the lamp and the wall is the sum of the distance \[u\] between the lamp and lens and the distance \[v\] between the image of lamp on wall and the lens.
\[d = u + v\]
Rearrange the above equation for \[v\].
\[v = d - u\]
Substitute \[6.0\,{\text{m}}\] for \[d\] and \[1.2\,{\text{m}}\] for \[u\] in the above equation.
\[v = \left( {6.0\,{\text{m}}} \right) - \left( {1.2\,{\text{m}}} \right)\]
\[ \Rightarrow v = 4.8\,{\text{m}}\]
Hence, the distance between the image of the lamp on the wall and the lens is \[4.8\,{\text{m}}\].

Let us now calculate the magnification of the image of the lamp.Since we have given that the image formed by the lens is real, we can conclude that the lens is a convex lens as the real image is formed only by the convex lens and the distance on front side of the convex lens are taken positive while the distance on back side of the convex lens are taken positive.
Substitute \[ - 1.2\,{\text{m}}\] for \[u\]and \[4.8\,{\text{m}}\] for \[v\] in equation (1).
\[m = - \dfrac{{4.8\,{\text{m}}}}{{ - 1.2\,{\text{m}}}}\]
\[ \therefore m = 4\]
Therefore, the magnification of the image of the lamp is 4.

Hence, the correct option is B.

Note:The students should not forget to take the distance of the lamp from the lens with a negative sign because the lens is a convex lens. If this negative sign is not used with the distance of the lamp from the lens, we will end with the magnification with the negative sign which means that the image of the lamp on the wall is diminished which is not the correct answer.