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Question

# A point object is placed at a distance of 15 cm from a convex lens. The image is formed on the other side at a distance of 30 cm from the lens. When a concave lens is placed in contact with the convex lens, the image shifts away further by 30 cm. Calculate the focal lengths of the two lenses.

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Solution

## Given, Distance between point object and convex lens, u = 15 cm Distance between the image of the point object and convex lens, v = 30 cm Let fc be the focal length of the convex lens. Then, using lens formula, we have: $\frac{1}{v}-\frac{1}{u}=\frac{1}{{f}_{c}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{f}_{\mathrm{c}}}=\frac{1}{30}-\frac{1}{\left(-15\right)}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{f}_{\mathrm{c}}}=\frac{1}{30}+\frac{1}{15}=\frac{3}{30}\phantom{\rule{0ex}{0ex}}⇒{f}_{\mathrm{c}}=10\mathrm{cm}$ Now, as per the question, the concave lens is placed in contact with the convex lens. So the image is shifted by a distance of 30 cm. Again, let vf be the final image distance from concave lens, then: ${v}_{\mathrm{f}}$ = + (30 + 30) = + 60 cm Object distance from the concave lens, v = 30 cm If fd is the focal length of concave lens then Using lens formula, we have: $\frac{1}{{v}_{\mathrm{f}}}-\frac{1}{v}=\frac{1}{{f}_{\mathrm{d}}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{f}_{\mathrm{d}}}=\frac{1}{60}-\frac{1}{30}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{f}_{d}}=\frac{30-60}{60×30}=\frac{-30}{60×30}\phantom{\rule{0ex}{0ex}}⇒{f}_{d}=-60\mathrm{cm}$ Hence, the focal length (fc ) of convex lens is 10 cm and that of the concave lens (fd ) is 60 cm.

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