Question

7. Focal power of the eye lens of a compound microscope is 6 dioptre. The microscope is to be
used to maximum magnifying power (angular magnification) of at least 12.5. The packing
instructions demand that length of the microscope should be 25 cm. Determine minimum
focal power of the objective. How much will its radius of curvature be if it is a biconvex lens
of n = 1.5.​

Answer 1

Answer:

Focal length of objective lens is 5 cm

Radius of curvature of the surface of lens is 5 cm

Explanation:

As we know that maximum magnification of compound microscope is given as

$M = \frac{L}{f_o}(1 + \frac{D}{f_e})$

now we know that

L = 25 cm

$f_e = \frac{1}{6} m$

D = 25 cm

M = 12.5

Now from above formula

$12.5 = \frac{25}{f_o}(1 + \frac{25}{100/6})$

$f_o = 2(1 + 1.5)$

$f_o = 5 cm$

Now from lens makers formula we have

$\frac{1}{f_o} = (n - 1)(\frac{2}{R})$

$\frac{1}{5} = (1.5 - 1)(\frac{2}{R})$

$R = 5 cm$

#Learn

Topic : Simple microscope

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