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Question: An object placed at a distance of 9cm from the first principal focus of a convex lens, produces a re...

An object placed at a distance of 9cm from the first principal focus of a convex lens, produces a real image at a distance of 25cm25cm from its second principal focus. Then focal length of the lens is:
A) 9cm9cm
B) 25cm25cm
C) 15cm15cm
D) 17cm17cm

Explanation

Solution

Using the mirror formula proceed with the given data. All the given data are in terms of focus, thus the object and image distance must be calculated in terms of focal length.

Complete step by step answer:
Let f be the focal length of the given convex lens.
We know, according to the new sign conventions, the following signs can be considered:
Since, we have a convex lens, the focal length lies on the left side of the lens and hence it is taken to be positive.
Object is always kept in the left hand side of the lens, thus object distance is also taken in negative.
Image is formed on the right hand side, thus has a positive sign.
As given in the question,
Let us consider:
u=u = Object Distance
v=v = Image Distance
As given in the question:
u=(9+f)u = - (9 + f)
v=+(25+f)v = + (25 + f)
Now, applying the Lens formula:
1f=1v1u\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}
Now, putting the values are mentioned above:
1f=1+(25+f)1(9+f)\dfrac{1}{f} = \dfrac{1}{{ + (25 + f)}} - \dfrac{1}{{ - (9 + f)}}
On solving the above equation, we obtain:
f2+25f+9f+225=2f2+34f{f^2} + 25f + 9f + 225 = 2{f^2} + 34f
From, here we obtain the unknown value:
f2=225cm{f^2} = 225cm
f=15cm\Rightarrow f = 15cm
Since, the focal length of a convex lens is positive. Thus:
f=15cmf = 15cm
This is our required answer.

Therefore, option (C) is correct.

Note: The signs of image distance, object distance and focal length must be considered according to the New Sign Convention. Considering the lens formula, if any two quantities from image distance, object distance and focal length if known, the other can be found.