Two convex lenses, each having a focal length of 10 cm , are positioned at a separation of 15 cm with their principal axes coinciding. Compute the focal length of the equivalent lens. (Note that the sign of the focal

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Geometrical Optics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Geometrical Optics

Two convex lenses, each having a focal length of 10 cm , are...

Two convex lenses, each having a focal length of

10 \text{ cm}

, are positioned at a separation of

15 \text{ cm}

with their principal axes coinciding. Compute the focal length of the equivalent lens. (Note that the sign of the focal length is positive even though the lens system actually diverges a parallel beam incident on it.)

Options

A $-5 \text{ cm}$
B $-20 \text{ cm}$
C $5 \text{ cm}$
D $20 \text{ cm}$

Correct answer

D. $20 \text{ cm}$

Step-by-step solution

Given

f_1 = 10 \text{ cm}

f_2 = 10 \text{ cm}

, and the distance between the lenses

d = 15 \text{ cm}

.

The equivalent focal length

F

of two thin lenses separated by a distance

d

is given by the formula:

\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2} - \dfrac{d}{f_1 f_2}

Substituting the given values:

\dfrac{1}{F} = \dfrac{1}{10} + \dfrac{1}{10} - \dfrac{15}{10 \times 10}

\dfrac{1}{F} = \dfrac{2}{10} - \dfrac{15}{100}

\dfrac{1}{F} = \dfrac{20}{100} - \dfrac{15}{100}

\dfrac{1}{F} = \dfrac{5}{100} = \dfrac{1}{20}

F = 20 \text{ cm}

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