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Ray Optics

Section: Physics

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Question : 6 of 17

Marks: +1, -0

Two thin convex lenses are kept in contact coaxially. If the focal length of the combination of the lenses is 4 cm and sum of the focal lengths of the two lenses is 18 cm , then the focal length of the lens of low power is

[AP EAPCET 21 May 2025 S2]

$8\text{ cm}$
$10\text{ cm}$
$6\text{ cm}$
$12\text{ cm}$

Solution:

f_1+f_2=18\text{ cm}

f=4\text{ cm}

\therefore \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}

\Rightarrow \frac{1}{4} = \frac{f_2+f_1}{f_1 f_2}

\Rightarrow \frac{1}{4} = \frac{18}{f_1 f_2}

\Rightarrow f_1 f_2 = 72

\Rightarrow f_1 (18-f_1) = 72

\Rightarrow 18 f_1 - f_1^2 = 72

\Rightarrow f_1^2 - 18 f_1 + 72 = 0

\Rightarrow f_1^2 - 12 f_1 - 6 f' + 72 = 0

\Rightarrow (f_1-12)(f_1-6) = 0

\therefore f_1=6\text{ cm},\ 12\text{ cm}

When,

f_1=6\text{ cm}

, then,

f_2=18-6=12\text{ cm}

When

f_1=12\text{ cm}

, then

f_2=18-12=6\text{ cm}

Since,

P \propto \frac{1}{f}

\therefore

Focal length of low power

=12\text{ cm}

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