Question

Calculate the radius of curvature of an equiconcave lens of refractive index $1.5$, when it is kept in a medium of refractive index $1.4$, to have power o f$-5D$?

Answer 1

Hint: Equiconcave lens can be defined as the lens that have the same curved surfaces from both sides and have the same focal length. In this question of optics, we have to calculate the radius of curvature of an equiconcave lens and for that we first calculate the focal length from the power given and then we will use Lens maker Formula in which we have to enter the value of focal length and refractive index. Also, we have to calculate the radii of curvature for equiconcave lens.

Complete answer:
As we have been given the value of power so, from this we will calculate the value of focal length by the below given formula –
$f=\dfrac {1} {P} $ where, f= focal length; P= power
$\begin {align}
  & \Rightarrow f=\dfrac {1} {-0.5} \\
 & \Rightarrow f=-0.2m \\
 & \therefore f=-20cm \\
\end{align}$
Now we will use Lens Maker Formula to calculate the radius of curvature –
The expression of the Lens Maker Formula is:
$\dfrac {1} {f} = (\mu -1)\left ( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$ where,
 $\mu $ = refractive index;
${{R} _ {1}}\, and\, {{R} _ {2}} $ = radii of curvature.
Here, the value of refractive index in Lens Maker Formula will be –
$\mu =\dfrac {{{\mu} _ {2}}} {{{\mu} _ {1}}} $ where,
${{\mu} _ {2}} $= refractive index of the equi concave lens,
${{\mu} _ {1}} $= refractive index of the medium.
So, the value of overall refractive index will be –
$\begin {align}
  & \Rightarrow \mu =\dfrac{1.5}{1.4} \\
 & \therefore \mu =1.0714 \\
\end{align}$
Now, as the lens is equiconcave so, one of its radius will be (–R) and the other will be (+R) i.e.
$\begin {align}
  & {{R} _ {1}} =-R \\
 & {{R} _ {2}} =+R \\
\end{align}$
Putting the values in Lens Maker Formula we get the expression as:
$\begin {align}
  & \Rightarrow \dfrac {1} {-20} = (1.0714-1) (\dfrac{-1}{R}-\dfrac{1}{R}) \\
 & \Rightarrow \dfrac {1} {-20} = (0.0714) (\dfrac{-2}{R}) \\
 & \Rightarrow R= (0.0714) (40) \\
 & \therefore R=2.857cm \\
\end{align}$
Therefore, the radius of curvature of equi concave lens will be $2.857cm$.

Note:
Always remember the sign conversion used in the lenses because most of us commit mistakes in it. To find the focal length keep in mind that it is reciprocal of power of lens for every such question. Convex lens is known as a converging lens and its focal length is always positive whereas, concave lens is known as a diverging lens and its focal length is always negative.