Question

The magnifications produced by a convex lens for two different positions of an object are m1 and m2 respectively (m1>m2) if "d" is the distance of seperation between the two positions of the object then focal length of the lens is
(1) √m1m2 (2) d/√m1-m2 (3) dm1m2/m1-m2 (4) d/m1-m2 correct answer is d/m1-m2 but how??

Answer 1

Answer on Question #71181, Physics / Optics

The magnifications produced by a convex lens for two different positions of an object are m1 and m2 respectively (m1>m2) if "d" is the distance of separation between the two positions of the object then focal length of the lens is

(1) Vm1m2 (2) d/Vm1-m2 (3) dm1m2/m1-m2 (4) d/m1-m2

correct answer is d/m1-m2 but how??

Solution:

Separation between object and image is

$$D = u + v$$

where $u$ and $v$ are the object and image distance.

So we have

$$m _ {1} = \frac {v}{u}$$

The second position of lens:

$$d = v - u$$

In this position:

$$m _ {2} = \frac {u}{v}$$

Thus,

$$m _ {1} m _ {2} = 1$$

From lens equation

$$\frac {1}{u} + \frac {1}{v} = \frac {1}{f}$$ $$v = m _ {1} u$$ $$\frac {1}{u} + \frac {1}{m _ {1} u} = \frac {1}{f}$$

So,

$$u = \frac {f (1 + m _ {1})}{m _ {1}}$$

From

$$u = m _ {2} v$$

we have

$$v = \frac {f (1 + m _ {2})}{m _ {2}}$$ $$d = v - u = f \left(\frac {1 + m _ {2}}{m _ {2}} - \frac {1 + m _ {1}}{m _ {1}}\right)$$ $$d = f \left(\frac {m _ {1} + m _ {1} m _ {2} - m _ {2} - m _ {1} m _ {2}}{m _ {2} m _ {1}}\right)$$

Thus, the focal length

$$f = \frac {d m _ {1} m _ {2}}{m _ {1} - m _ {2}}$$

In our case

$$m _ {1} m _ {2} = 1$$ $$f = \frac {d}{m _ {1} - m _ {2}}$$

Answer: $f = \frac{d}{m_1 - m_2}$

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