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Question: A lens of refractive index n is put in a liquid of refractive index n’ of focal length of lens in ai...

A lens of refractive index n is put in a liquid of refractive index n’ of focal length of lens in air is f, its focal length in liquid will be
a)fn(n1)(nn)-\dfrac{fn'\left( n-1 \right)}{\left( n'-n \right)}
b)f(nn)n(n1)-\dfrac{f\left( n'-n \right)}{n'\left( n-1 \right)}
c)n(n1)f(nn)-\dfrac{n'\left( n-1 \right)}{f\left( n'-n \right)}
d)fnn(nn)\dfrac{fn'n}{\left( n'-n \right)}

Explanation

Solution

Lens maker’s formula 1f=(n1)(1R11R2)\dfrac{1}{f}=\left( n-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)where, f = the focal length, R2{{R}_{2}} and R2{{R}_{2}} are the radius of curvatures of the lens.

Complete step by step answer:
By using the lens maker’s formula for air as a medium, we have:
1f=(n1)(1R11R2)......(1)\dfrac{1}{f}=\left( n-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)......(1)
Now, it is given that n = 1 is the refractive index in air and,
n`= refractive index in liquid which is another medium
Putting, the refractive indices of two different media we get,
1fl=(aμl1)(1R11R2)......(2)\dfrac{1}{{{f}_{l}}}=\left( ^{a}{{\mu }_{l}}-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)......(2)
where, aμl=μaμl^{a}{{\mu }_{l}}=\dfrac{{{\mu }_{a}}}{{{\mu }_{l}}}
As we have:
μa=n μl=n \begin{aligned}& {{\mu }_{a}}=n \\\ & {{\mu }_{l}}=n' \\\ \end{aligned}
So, we gave:
aμl=nn......(3)^{a}{{\mu }_{l}}=\dfrac{n}{n'}......(3)
Now, put the value of aμl^{a}{{\mu }_{l}} from equation (3) in equation (2, we get:
1fl=(nn1)(1R11R2)......(4)\dfrac{1}{{{f}_{l}}}=\left( \dfrac{n}{n'}-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)......(4)

where, fl{{f}_{l}} focal length of lens in liquid

Now, divide equation (1) by equation (4), we get:

1f1fl=(n1)(1R11R2)(nn1)(1R11R2) \dfrac{\dfrac{1}{f}}{\dfrac{1}{{{f}_{l}}}}=\dfrac{\left( n-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)}{\left( \dfrac{n}{n'}-1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)} \\\
    flf=(n1)(nn1)\implies \dfrac{{{f}_{l}}}{f}=\dfrac{\left( n-1 \right)}{\left( \dfrac{n}{n'}-1 \right)}
    flf=n(n1)(nn)......(5)\implies \dfrac{{{f}_{l}}}{f}=\dfrac{n'\left( n-1 \right)}{\left( n-n' \right)}......(5)
So, we have:
{{f}_{l}}=\dfrac{n'\left( n-1 \right)f}{\left( n-n' \right)} \\\
=fn(nn)n(n1)=-\dfrac{fn'\left( n'-n \right)}{n'\left( n-1 \right)}

So, the correct answer is “Option A”.

Additional Information:
Lens maker’s formula is used by the manufacturers to make the lens of a particular power from the glass of a given refractive index.
It has certain limitations
The lenses need to be thin because the separation between the two refracting surfaces will be small.
The medium on both sides of the lens need to be the same.

Note:
Sign convention
For convex lenses, focal length is positive and for concave lenses the focal length is negative.
We must follow the sign convention to correctly solve any type of problem related to lens maker’s formula.