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Double convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20cm?  
A. 11cm
B. 22cm
C. 7cm
D. 6cm

Answer
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Hint: The focal length of the lens depends on the refractive index of the lens and the radii of curvatures of the spherical surfaces on both the sides such that $\dfrac{1}{f}=\left( \mu -1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$. Use this formula to find the radius of curvature.

Complete step by step answer:
A double convex lens is a lens that has convex surfaces on both the sides. The focal length of a lens is the distance between the focus of the lens and the optical centre of the lens. For a convex lens, its focus is a point where all the parallel rays of light that are also parallel to the optical axis intersect at a point after passing through the lens. The point of focus is always on the optical axis.
Suppose the lens is placed in air. The focal length of the lens depends on the refractive index of the lens and the radii of curvatures of the spherical surfaces on both the sides such that $\dfrac{1}{f}=\left( \mu -1 \right)\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$…..(i), (according to the sign convection). 
This formula is called the lens maker’s formula.
Here, f is the focal length of the lens, $\mu $ is the refractive index of the material from which the lens is made. ${{R}_{1}}$ is the radius of the spherical surface on the which the light rays are incident and ${{R}_{2}}$ is the radius of the spherical surface on the other side, according to the sign convection.
It is given that the focal length of the lens is 20cm.
f=20cm.
It is given that the lens is made from glass and its refractive index is 1.55.
$\mu $=1.55.
It is given that the radii of the spherical surfaces are the same. According to sign convection, ${{R}_{1}}=R$ and ${{R}_{2}}=-R$.
Substitute the values in equation (i).
$\dfrac{1}{20}=\left( 1.55-1 \right)\left( \dfrac{1}{R}-\dfrac{1}{(-R)} \right)$
$\Rightarrow \dfrac{1}{20}=\left( 0.55 \right)\left( \dfrac{1}{R}+\dfrac{1}{R} \right)$
$\Rightarrow \dfrac{1}{20\left( 0.55 \right)}=\left( \dfrac{2}{R} \right)$
$\therefore R=2\times 20\times 0.55=22cm$
Therefore, the radius of curvature required to make the given lens is 22cm.

Note: Students may make mistakes in using the sign convection.
They may also make mistakes in selecting the correct ${{R}_{1}}$ and ${{R}_{2}}$. 
Note that the lens maker’s formula is only applicable for thin lenses. This means that the thickness of the lens must be very small compared to the focal length of the lens.