An object is placed 15 cm in front of a convex lens of focal length 10 cm. Find the nature and position of the image formed. Where should a concave mirror of radius of curvature 20 cm be placed so that the final image is formed at the position of the object itself?
CBSE OUTSIDE DELHI - SET 1 PANCHKULA 2015
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For lens $\frac{1}{\text{v}} - \frac{1}{\text{u}} = \frac{1}{\text{f}}$
$\frac{1}{\text{v}} - \frac{1}{-15} = \frac{1}{+10}$
$\frac{1}{\text{v}} + \frac{1}{15} = \frac{1}{10}$
$\therefore\text{v} = 30\text{cm}$
Nature of image- real, magnified
Final image formed will be at the object itself only if image formed by lens is at the position of centre of curvature of mirror.
$\therefore\text{D} = (30 + \text{R})\text{cm} =(30 +20)\text{cm} = 50 \text{cm}.$
(Distance of mirror from lens).
$\frac{1}{\text{v}} - \frac{1}{-15} = \frac{1}{+10}$
$\frac{1}{\text{v}} + \frac{1}{15} = \frac{1}{10}$
$\therefore\text{v} = 30\text{cm}$
Nature of image- real, magnified
Final image formed will be at the object itself only if image formed by lens is at the position of centre of curvature of mirror.
$\therefore\text{D} = (30 + \text{R})\text{cm} =(30 +20)\text{cm} = 50 \text{cm}.$
(Distance of mirror from lens).
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