Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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

Last updated date: 13th Jun 2024
Total views: 374.1k
Views today: 9.74k
Verified
374.1k+ views
Hint: Convex lens: It is converging lens. Lens maker’s formula for convex lens:
$\dfrac{1}{\text{f}}=\left( \text{u}-1 \right)\left( \dfrac{1}{{{\text{R}}_{1}}}-\dfrac{1}{{{\text{R}}_{2}}} \right)$
Where u = refractive index
${{\text{R}}_{1}}=$ radius of curvature for 1 sphere
${{\text{R}}_{2}}=$ radius of curvature for 1 sphere
f = focal length.

Complete step by step solution
The radii of curvature of two surfaces at convex lens
\begin{align} & {{\text{R}}_{1}}=0\cdot 2\text{ m} \\ & {{\text{R}}_{2}}=-0\cdot 22 \\ & {{\text{n}}_{\text{air}}}=1\cdot 5 \\ \end{align}
In air
\begin{align} & \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{align}
When lens is immersed in water
${{\text{u}}^{1}}=\dfrac{{{\text{u}}_{\text{air}}}}{{{\text{u}}_{\text{water}}}}=\dfrac{1\cdot 5}{1\cdot 35}=1\cdot 1278$
Now,
\begin{align} & \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{align}
Hence, when lens is immersed in water focal length of lens is increased by
\begin{align} & =\left( 0\cdot 82-0\cdot 209 \right) \\ & =0\cdot 611\text{ m} \\ \end{align}

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