When the two lens of focal length f1 and f2 are kept in contact then focal length of the combination is?

Option 3 : \(\dfrac{f_1 f_2}{f_1 + f_2}\)

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CONCEPT:

• The focal length of the combination: When two lenses of focal length f1 and f2 are placed in contact, the effective focal length f of the combination is given by:

\( {1 \over f}= {1 \over f_1} +{1 \over f_2}\)

​CALCULATION:

Given that Two thin lenses are of focal lengths f1 and f2 

The focal length of the combination \( {1 \over f}= {1 \over f_1} +{1 \over f_2}\)

\(f=\dfrac{f_1 f_2}{f_1 + f_2}\)

So the correct answer is option 3.

• The focal length of the convex lens is positive and that of the concave lens is negative. 

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Text Solution

Solution : If two lenses of focal lengths, `F_(1)` and `f_(2)` are kept in contact with each other , the focal length (f) of the combination is given by `1/f =(1)/(f_(2)) +(1)/(f_(2))`/ <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NVT_SCI_TECH_DIG_X_P1_C07_E10_045_S01.png" width="80%"> <br> If `P_(1) and P_(2)` are the powers of these lenses, the power (P) of the combination is given by `P=P_(1)+P_(2)`

Let two thin lenses L1 and L2 of focal lengths f1 and f2 be put in contact. O is a point object at a distance u from the lens L1 Its image is formed at I after refraction through the two lenses at a distance v from the combination. The lens L1 forms the image of O at I’. I’, then serves as a virtual object for the lens L2 which forms a real image at I.

Now, we know that by lens formula
`1/v - 1/u = 1/f`

Applying this relation for refraction at the lens
L1 of focal length f1 can be written as

`1/v' - 1/u = 1/f_1`    ...(i)

For refraction at the lens L2
u = v', v = v

∴ For this lens `1/v - 1/v' = 1/f_2`    ...(ii)

And if F is the focal length of the combination, then

`1/v - 1/u = 1/F`

Adding (i) and (ii), we get

`1/v' - 1/u + 1/v - 1/v' = 1/f_1  + 1/f_2`

or `1/v -1/u = 1/f_1  + 1/f_2`

or `1/"F" = 1/f_1  + 1/f_2 `