Question

A rod length 10cm lies along the principal axis of a concave mirror of focal length \[10\,{\text{cm}}\] in such a way that its end closer to the pole is 20cm away from the mirror. The length of image is:
A.\[2.5\,{\text{cm}}\]
B.\[5\,{\text{cm}}\]
C.\[10\,{\text{cm}}\]
D.\[15\,{\text{cm}}\]

Answer 1

Hint: Use the mirror formula. This formula gives the relation between the focal length of mirror, object distance from mirror and image distance from mirror. Determine the distances of the two ends of the rod from the concave mirror and the use mirror formula to determine the distances of two ends of the image of the rod from the mirror.

Formula used:
The mirror formula is given by
\[\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}\] …… (1)
Here, \[f\] is the focal length of the mirror, \[u\] is the distance of the object from the mirror and \[v\] is the distance of image from the mirror.

Complete step by step solution:
We have given that the length of the rod is \[10\,{\text{cm}}\]. The end of the rod closer to the pole is \[20\,{\text{cm}}\] away from the concave mirror.The focal length of the concave mirror is \[10\,{\text{cm}}\].
\[f = 10\,{\text{cm}}\]
Let AB be the length of the rod.Draw the diagram of the concave mirror showing the distances of the ends of the rod from the concave mirror.

From the above diagram, it can be concluded that the distance \[{u_A}\] of end A of the rod is \[20\,{\text{cm}}\] away from the mirror and the distance \[{u_B}\] of and B of the rod is \[30\,{\text{cm}}\] away from the mirror.
We have to find the length of the image.
For this, we need to determine the distance \[{v_A}\] of the end A of the image of the rod and the distance \[{v_B}\] of the end B of the image from the rod.
Rewrite equation (1) for the image distance of end A.
\[\dfrac{1}{f} = \dfrac{1}{{{u_A}}} + \dfrac{1}{{{v_A}}}\]
Substitute \[ - 20\,{\text{cm}}\] for \[{u_A}\] and \[ - 10\,{\text{cm}}\] for \[f\] in the above equation.
\[\dfrac{1}{{ - 10\,{\text{cm}}}} = \dfrac{1}{{ - 20\,{\text{cm}}}} + \dfrac{1}{{{v_A}}}\]
\[ \Rightarrow \dfrac{1}{{{v_A}}} = \dfrac{1}{{20}} - \dfrac{1}{{10}}\]
\[ \Rightarrow \dfrac{1}{{{v_A}}} = \dfrac{{10 - 20}}{{200}}\]
\[ \Rightarrow \dfrac{1}{{{v_A}}} = \dfrac{{ - 10}}{{200}}\]
\[ \Rightarrow {v_A} = - 20\,{\text{cm}}\]
Hence, the distance of end A of the image of the rod from the concave mirror is \[ - 20\,{\text{cm}}\].
Rewrite equation (1) for the image distance of end B.
\[\dfrac{1}{f} = \dfrac{1}{{{u_B}}} + \dfrac{1}{{{v_B}}}\]
Substitute \[ - 30\,{\text{cm}}\] for \[{u_B}\] and \[ - 10\,{\text{cm}}\] for \[f\] in the above equation.
\[\dfrac{1}{{ - 10\,{\text{cm}}}} = \dfrac{1}{{ - 30\,{\text{cm}}}} + \dfrac{1}{{{v_B}}}\]
\[ \Rightarrow \dfrac{1}{{{v_B}}} = \dfrac{1}{{30}} - \dfrac{1}{{10}}\]
\[ \Rightarrow \dfrac{1}{{{v_B}}} = \dfrac{{10 - 30}}{{300}}\]
\[ \Rightarrow \dfrac{1}{{{v_B}}} = \dfrac{{ - 20}}{{300}}\]
\[ \Rightarrow {v_B} = - 15\,{\text{cm}}\]
Hence, the distance of end B of the image of the rod from the concave mirror is \[ - 15\,{\text{cm}}\].
The length of the image of the rod is given by
\[L = {v_B} - {v_A}\]
Substitute \[ - 15\,{\text{cm}}\] for \[{v_B}\] and \[ - 20\,{\text{cm}}\] for \[{v_A}\] in the above equation.
\[L = \left( { - 15\,{\text{cm}}} \right) - \left( { - 20\,{\text{cm}}} \right)\]
\[ \therefore L = 5\,{\text{cm}}\]
Therefore, the length of the image of the rod is \[5\,{\text{cm}}\].

Hence, the correct option is B.

Note:The students may think why the values of the focal length and object distances are taken negative while substituting in the formula. For the concave mirror, the distances measured in front of the mirror are taken negative and the distance measured behind the mirror are taken positive.