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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:
\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f_c}\]
\[ \Rightarrow \frac{1}{f_c} = \frac{1}{30} - \frac{1}{( - 15)}\]
\[ \Rightarrow \frac{1}{f_c} = \frac{1}{30} + \frac{1}{15} = \frac{3}{30}\]
\[ \Rightarrow f_c = 10 \text{ cm }\]
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:
\[v_f\] = + (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:
\[\frac{1}{v_f} - \frac{1}{v} = \frac{1}{f_d}\]
\[ \Rightarrow \frac{1}{f_d} = \frac{1}{60} - \frac{1}{30}\]
\[ \Rightarrow \frac{1}{f_d} = \frac{30 - 60}{60 \times 30} = \frac{- 30}{60 \times 30}\]
\[ \Rightarrow f_d = - 60 \text{ cm }\]
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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