Respuesta :
Given:
The power of the lens is,
[tex]P=+1.5\text{ D}[/tex]The refractive index of the lens is,
[tex]n=1.6[/tex]The radius of curvature of the convex surface is,
[tex]\begin{gathered} R_1=20\text{ cm} \\ =0.20\text{ m} \end{gathered}[/tex]To find:
The radius of curvature of the other surface
Explanation:
If the radius of curvature of the other surface is
[tex]R_2[/tex]we can write,
[tex]\frac{1}{f}=P=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})[/tex]Now, substituting the values we get,
[tex]\begin{gathered} 1.5=(1.6-1)(\frac{1}{0.20}-\frac{1}{R_2}) \\ \frac{1.5}{0.6}=\frac{1}{0.20}-\frac{1}{R_2} \\ \frac{1}{R_2}=\frac{1}{0.20}-\frac{1.5}{0.6} \\ \frac{1}{R_2}=2.5 \\ R_2=0.40\text{ m} \\ R_2=40\text{ cm} \end{gathered}[/tex]Hence, the radius of curvature of the other surface is 40 cm.
