Question

Use the mirror equation to show that an object placed between f and 2f of a concave mirror forms an image beyond 2f.

Answer 1

\[\text { For a concave mirror, the focal length (f) is negative } . \]

\[ \therefore f < 0\]

\[\text { When the object is placed on the left side of the mirror, the object distance (u) is negative } . \]

\[ \therefore u < 0\]

\[\text { For image distance v, we can write the mirror formula as }: \]

\[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\]

\[\frac{1}{v} = \frac{1}{f} - \frac{1}{u} . . . . . (i) \]

\[\text { The object lies between f and 2f }\]

\[ \Rightarrow 2f < u < f ( \because \text { u and f are negative })\]

\[ \frac{1}{2f} > \frac{1}{u} > \frac{1}{f}\]

\[ - \frac{1}{2f} < - \frac{1}{u} < - \frac{1}{f}\]

\[\frac{1}{f} - \frac{1}{2f} < \frac{1}{f} - \frac{1}{u} < 0\]

\[\text { Using equation (i), we get }: \]

\[\frac{1}{2f} < \frac{1}{v} < 0\]

\[ \therefore \frac{1}{v}\text {  is negative, i . e . , v is negative }\]

\[\frac{1}{2f} < \frac{1}{v}\]

\[ 2f > v\]

Therefore, the image lies beyond 2f.

\[ - v > - 2f\]