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Question: The radii of curvature of two surfaces of convex lens is \( 0\cdot 2\text{ m} \) and \( 0\cdot 22\te...

The radii of curvature of two surfaces of convex lens is 0⋅2 m0\cdot 2\text{ m} and 0⋅22 m0\cdot 22\text{ m} . Find the length of the lens if the refractive index of the focal material of the lens is 1⋅5 m1\cdot 5\text{ m} . Also, find the charge in focal length, if it is immersed in H2O{{\text{H}}_{2}}\text{O} of refractive index 1⋅331\cdot 33 .

Explanation

Solution

Convex lens: It is converging lens. Lens maker’s formula for convex lens:
1f=(u−1)(1R1−1R2)\dfrac{1}{\text{f}}=\left( \text{u}-1 \right)\left( \dfrac{1}{{{\text{R}}_{1}}}-\dfrac{1}{{{\text{R}}_{2}}} \right)
Where u = refractive index
R1={{\text{R}}_{1}}= radius of curvature for 1 sphere
R2={{\text{R}}_{2}}= radius of curvature for 1 sphere
f = focal length.

Complete step by step solution
The radii of curvature of two surfaces at convex lens
R1=0⋅2 m R2=−0⋅22 nair=1⋅5 \begin{aligned} & {{\text{R}}_{1}}=0\cdot 2\text{ m} \\\ & {{\text{R}}_{2}}=-0\cdot 22 \\\ & {{\text{n}}_{\text{air}}}=1\cdot 5 \\\ \end{aligned}
In air
1f=(n−1)(1R1−1R2) =(1⋅5−1)(10⋅2+10⋅22) =0⋅5[5+4⋅54]=0⋅5×9⋅54 =4⋅77 f=14⋅77  =0⋅2096 m \begin{aligned} & \dfrac{1}{\text{f}}=\left( \text{n}-1 \right)\left( \dfrac{1}{{{\text{R}}_{1}}}-\dfrac{1}{{{\text{R}}_{2}}} \right) \\\ & =\left( 1\cdot 5-1 \right)\left( \dfrac{1}{0\cdot 2}+\dfrac{1}{0\cdot 22} \right) \\\ & =0\cdot 5\left[ 5+4\cdot 54 \right]=0\cdot 5\times 9\cdot 54 \\\ & =4\cdot 77 \\\ & \text{f}=\dfrac{1}{4\cdot 77} \\\ & \text{ }=0\cdot 2096\text{ m} \\\ \end{aligned}
When lens is immersed in water
u1=uairuwater=1⋅51⋅35=1⋅1278{{\text{u}}^{1}}=\dfrac{{{\text{u}}_{\text{air}}}}{{{\text{u}}_{\text{water}}}}=\dfrac{1\cdot 5}{1\cdot 35}=1\cdot 1278
Now,
1f1=(u1−1)(1R1−1R2) =(1⋅1278−1)(10⋅2−1−0⋅22) =0⋅1278(5+4⋅54) =1⋅219 f1=11⋅1219=0⋅82 m \begin{aligned} & \dfrac{1}{{{\text{f}}^{1}}}=\left( {{\text{u}}^{1}}-1 \right)\left( \dfrac{1}{{{\text{R}}_{1}}}-\dfrac{1}{{{\text{R}}_{2}}} \right) \\\ & =\left( 1\cdot 1278-1 \right)\left( \dfrac{1}{0\cdot 2}-\dfrac{1}{-0\cdot 22} \right) \\\ & =0\cdot 1278\left( 5+4\cdot 54 \right) \\\ & =1\cdot 219 \\\ & {{\text{f}}^{1}}=\dfrac{1}{1\cdot 1219}=0\cdot 82\text{ m} \\\ \end{aligned}
Hence, when lens is immersed in water focal length of lens is increased by
=(0⋅82−0⋅209) =0⋅611 m \begin{aligned} & =\left( 0\cdot 82-0\cdot 209 \right) \\\ & =0\cdot 611\text{ m} \\\ \end{aligned}

Note
We can use the lens maker’s formula for finding the focal length of the lens. We can find the radius of curvature and refractive index also from the lens maker’s formula. While solving numerical keep in mind that all values should be in S.I. units