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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.
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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