Answer:
The radii of curvature of its surface is 0.00212 cm and after reduced the radii of curvature of its surface is 0.002 cm.
Explanation:
Given that,
Refractive index = 1.530 cm
After reduce refractive index = 1.500
Focal length =250.0 cm
Radius of curvature of plane mirror [tex]r_{2}=\infty[/tex]
We need to calculate the radii of curvature of lens
Using formula of lens
[tex]f=(n-1)({\dfrac{1}{r_{1}}-\dfrac{1}{r_{2}}})[/tex]
Put the value into the formula
[tex]250.0=(1.530-1)({\dfrac{1}{r_{1}}-\dfrac{1}{\infty}})[/tex]
[tex]\dfrac{1}{r_{1}}=\dfrac{250.0}{1.530-1}[/tex]
[tex]r_{1}=0.00212\ cm[/tex]
If n was reduced to 1.500
Then, the radii of curvature of lens
[tex]250.0=(1.500-1)({\dfrac{1}{r_{1}}-\dfrac{1}{\infty}})[/tex]
[tex]\dfrac{1}{r_{1}}=\dfrac{250.0}{1.500-1}[/tex]
[tex]r_{1}=0.002\ cm[/tex]
Hence, The radii of curvature of its surface is 0.00212 cm and after reduced the radii of curvature of its surface is 0.002 cm.
