Geometrical OpticsMedium
Question
A rod of length 10 cm lies along the principle axis of a concave mirror of focal length 10 cm as shown in figure. The length of the image is $\_\_\_\_$ cm.
Options
A.2.5
B.5
C.7.5
D.7
Solution
Sol. [Key Concept: Apply the mirror formula at both ends of the rod; image length $= {|v}_{1} - v_{2}|$.]
For A $u = - 20\text{ }cm,f = - 10\text{ }cm$
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\ \therefore v = \frac{u \times f}{u - f} = - 20\text{ }cm$$
$$\boxed{\begin{matrix} \text{~}\text{For}\text{~}B & u = - 30\text{ }cm,f = - 10\text{ }cm \\ & \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ & v = \frac{uf}{u - f} = - 15\text{ }cm \end{matrix}}$$
Length of image $= 20 - 15 = 5\text{ }cm$
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