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प्रश्न
Use the mirror equation to show that an object placed between f and 2f of a concave mirror forms an image beyond 2f.
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उत्तर
\[\text { For a concave mirror, the focal length (f) is negative } . \]
\[ \therefore f < 0\]
\[\text { When the object is placed on the left side of the mirror, the object distance (u) is negative } . \]
\[ \therefore u < 0\]
\[\text { For image distance v, we can write the mirror formula as }: \]
\[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\]
\[\frac{1}{v} = \frac{1}{f} - \frac{1}{u} . . . . . (i) \]
\[\text { The object lies between f and 2f }\]
\[ \Rightarrow 2f < u < f ( \because \text { u and f are negative })\]
\[ \frac{1}{2f} > \frac{1}{u} > \frac{1}{f}\]
\[ - \frac{1}{2f} < - \frac{1}{u} < - \frac{1}{f}\]
\[\frac{1}{f} - \frac{1}{2f} < \frac{1}{f} - \frac{1}{u} < 0\]
\[\text { Using equation (i), we get }: \]
\[\frac{1}{2f} < \frac{1}{v} < 0\]
\[ \therefore \frac{1}{v}\text { is negative, i . e . , v is negative }\]
\[\frac{1}{2f} < \frac{1}{v}\]
\[ 2f > v\]
Therefore, the image lies beyond 2f.
\[ - v > - 2f\]
