Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

A small linear object of length \[b\] is placed at a distance \[u\] from the pole of the concave mirror along the axis away from the mirror. The focal length of the mirror is \[f\]. The length of the image will then be?
A. \[b{\left( {\dfrac{f}{{u - f}}} \right)^2}{\left( {\dfrac{f}{{u - f + b}}} \right)^2}\]
B. \[b{\left( {\dfrac{f}{{u - f}}} \right)^2}\left( {\dfrac{f}{{u - f + b}}} \right)\]
C. \[b\left( {\dfrac{f}{{u - f}}} \right){\left( {\dfrac{f}{{u - f + b}}} \right)^2}\]
D. \[b\left( {\dfrac{f}{{u - f}}} \right)\left( {\dfrac{f}{{u - f + b}}} \right)\]

Answer
VerifiedVerified
553.2k+ views
Hint:We will use the mirror formula, to find out the image distance at the near point. We will again make use of the mirror formula to find out the image distance at the far point. After that the difference of these distances will give us the length of the image.

Formula used:
Using mirror formula,
\[\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}\] …… (1)
Where,
\[u\] is distance of the object
\[v\] is distance of the image
\[f\] is the focal length of the mirror.

Complete step by step answer:
Given that, \[b\] is the object placed at a distance \[u\] from the pole of the concave mirror along the axis away from the mirror.
Here, \[{u_A} = - u\] and focal length \[ - f\].
seo images

Using mirror formula, we get,
$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f} \\
\Rightarrow \dfrac{1}{{ - u}} + \dfrac{1}{{{v_A}}} = \dfrac{{ - 1}}{f} \\
\Rightarrow {v_A} = - \left( {\dfrac{{uf}}{{u - f}}} \right) \\$
Again, for end B,
\[{u_B} = - \left( {u + b} \right)\],

Putting the value, we get,
$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f} \\
\Rightarrow \dfrac{1}{{ - \left( {u + b} \right)}} + \dfrac{1}{{{v_B}}} = \dfrac{1}{{ - f}} \\
\Rightarrow \dfrac{1}{{{v_B}}} = \dfrac{1}{{ - f}} + \dfrac{1}{{\left( {u + b} \right)}} \\
\Rightarrow {v_B} = \dfrac{{\left( {u + b} \right)f}}{{\left( {u + b} \right) - f}} \\$
So, the length of the image, \[{v_A} - {v_B}\]
$\Rightarrow {v_A} - {v_B} = \dfrac{{ - uf}}{{u - f}} + \dfrac{{\left( {u + b} \right)f}}{{\left( {u + b} \right) - f}} \\
\Rightarrow{v_A} - {v_B} = \dfrac{f}{{u - f}}\left( {\dfrac{{u + b}}{{1 + \dfrac{b}{{u - f}}}} - u} \right) \\
\Rightarrow{v_A} - {v_B} = \dfrac{f}{{u - f}}\left( {\dfrac{{u + b - u - \dfrac{{bu}}{{u - f}}}}{{1 + \dfrac{b}{{u - f}}}}} \right) \\
\Rightarrow{v_A} - {v_B} = b\left( {\dfrac{f}{{u - f}}} \right)\left( {\dfrac{{1 - \dfrac{u}{{u - f}}}}{{1 + \dfrac{b}{{u - f}}}}} \right) \\
\Rightarrow{v_A} - {v_B} = b\left( {\dfrac{f}{{u - f}}} \right)\left( {\dfrac{{\dfrac{{ - f}}{{u - f}}}}{{\dfrac{{u - f + b}}{{u - f}}}}} \right) \\
\therefore{v_A} - {v_B} = b\left( {\dfrac{f}{{u - f}}} \right)\left( {\dfrac{{ - f}}{{u - f + b}}} \right) \\$
Hence, the required answer is \[b\left( {\dfrac{f}{{u - f}}} \right)\left( {\dfrac{f}{{u - f + b}}} \right)\].

The correct option is D.

Additional information:
Mirror formula: One of the most common questions posed in different board examinations as well as competitive examinations is the derivation of the mirror formula. A mirror formula can be defined as the formula that connects the distance between the ‘ \[u\] ’ object, the ‘ \[v\] ’ image distance, and the ‘ \[f\] ’ mirror focal length. For both plane mirrors and spherical mirrors, the mirror formula applies to (convex and concave mirrors). Here the derivation of the mirror formula is given so that students can better understand the definition of the subject. The formula for the mirror is written as: \[\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\].

Note:Remember that the lens formula is \[{\text{1/f}} = {\text{1/v}} - {\text{1/u}}\] and the mirror formula is \[{\text{1/f}} = {\text{1/v}} + {\text{1/u}}\]. The optical core of the lens is considered for measuring all the distances. They are known to be negative when the distances are determined opposite to the direction of the incident light. They are assumed to be positive when the distances are determined in the same direction as the incident light. They are known to be positive when the heights are measured upwards and perpendicular to the principal axis. They are known to be negative when the heights are measured downwards and perpendicular to the principal axis.