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A biconvex lens of refractive index 1.5 has a focal length of 20 cm in air. Its focal length when immersed in a liquid of refractive index 1.6 will be

Question

A biconvex lens of refractive index 1.5 has a focal length of 20 cm in air. Its focal length when immersed in a liquid of refractive index 1.6 will be

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Solution

The focal length of a lens is given by the lensmaker's formula:

1/f = (n-1)(1/R1 - 1/R2)

where: f is the focal length, n is the refractive index of the lens, R1 and R2 are the radii of curvature of the two lens surfaces.

Given that the lens is biconvex, R1 = -R2. Therefore, the formula simplifies to:

1/f = 2(n-1)/R

When the lens is in air (n=1.5), the focal length f is 20 cm. So we can write:

1/20 = 2(1.5-1)/R => R = 2(0.5)/0.05 = 20 cm

Now, when the lens is immersed in a liquid of refractive index 1.6, we need to find the new focal length f'. We can use the same formula, but now n=1.6 and we know R=20 cm:

1/f' = 2(1.6-1)/20 => f' = 1/(2*0.6/20) = 1/(0.06) = 16.67 cm

So, the focal length of the lens when immersed in the liquid will be approximately 16.67 cm.

This problem has been solved

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