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Question: Determine the focal length of convex lens on the following points by describing the displacement met...

Determine the focal length of convex lens on the following points by describing the displacement method:
i) Derivation of Formula
ii)Draw a ray diagram
iii)Observation table
iv) Precautions (any 2)

Explanation

Solution

Let there be two pins placed. Then, find out the distance of the object and then find the distance of the image. Now, use the lens equation,1f=1v1u\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}
Now, put the values of distance of image and object respectively in the above lens equation. From this, find out the focal length of the lens.

Complete step by step Solution:
(i) Formula Derivation: Let there be two pins ABAB and CDCDrespectively placed such that the distance between both of them is more than ufuf.
When the lens is at LLthen, let the image formed be ABA'B'on pin CDCD
OA=u OA=v d=v+u(1)  \therefore OA = u \\\ OA' = v \\\ d = v + u \cdots (1) \\\
The image is again formed on CDCD because AA and AA' are conjugate foci. This happens in the second position of L2{L_2}
OA=u OA=v  O'A' = u \\\ O'A = v \\\
Let the displacement of the lens be xx. Therefore,
x=vu(2)x = v - u \cdots (2)
Using the value of vv from equation (1)(1)in equation (2)(2)we get,
u=dx2u = \dfrac{{d - x}}{2}
Similarly, using the value of uu from equation (1)(1)in equation (2)(2)we get,
v=d+x2v = \dfrac{{d + x}}{2}
We know that, the lens formula is
1f=1v1u\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}
Where, f=f = focal length, u=u = object distance and v=v = image distance
Putting the values of vvand uuin lens formula
1f=1d+x21dx2\dfrac{1}{f} = \dfrac{1}{{\dfrac{{d + x}}{2}}} - \dfrac{1}{{\dfrac{{d - x}}{2}}}
According to the sign convention, uuwill be positive and vvis negative.
1f=1(d+x)21(dx)2 1d+x2+1dx2 1f=2d+x+2dx 2d2x+2d+2x(d+x)(dx) 1f=4dd2x2 f=d2x24d  \dfrac{1}{f} = \dfrac{1}{{\dfrac{{\left( {d + x} \right)}}{2}}} - \dfrac{1}{{ - \dfrac{{\left( {d - x} \right)}}{2}}} \\\ \Rightarrow \dfrac{1}{{\dfrac{{d + x}}{2}}} + \dfrac{1}{{\dfrac{{d - x}}{2}}} \\\ \dfrac{1}{f} = \dfrac{2}{{d + x}} + \dfrac{2}{{d - x}} \\\ \Rightarrow \dfrac{{2d - 2x + 2d + 2x}}{{(d + x)(d - x)}} \\\ \dfrac{1}{f} = \dfrac{{4d}}{{{d^2} - {x^2}}} \\\ f = \dfrac{{{d^2} - {x^2}}}{{4d}} \\\
This is the required formula.

(ii) Ray diagram: The ray diagram can be made as

(iii) Observation Table

Position of object pinPosition of image pin1st position of lens2nd position of lensDistance between pinsDisplacement of lensff
acma cmbcmb cmmcmm cmncmn cmd=bad = b - ax=nmx = n - md2x2udcm\dfrac{{{d^2} - {x^2}}}{{ud}}cm

(iv)Precautions: The line which joins the tip of pins and optical centre should lie on a horizontal line.
The distance between pins should be greater than 4 times of focal length.

Note: The focal length of lens should be four times less than the distance between two pins.
When using the lens formula, keep in mind the sign convention of distance of object, distance of image and focal length.