Question
Lenses are constructed by a material of refractive index $1.50$. The magnitude of the radii of curvature are $20cm$ and $30cm$. Find the focal lengths of the possible lenses with the above specifications.
✓
Answer
Given $\mu=1.5$ Magnitude of radii of curvatures = 20cm and 30cm The 4types of possible lens are as below.$\frac{1}{\text{f}}=(\mu-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$
Case (1): (Double convex) $[R_1 = +ve, R_2 = -ve]\frac{1}{\text{f}}=(15-1)\Big(\frac{1}{20}-\frac{1}{30}\Big)\Rightarrow\text{f}=24\text{cm}$
Case (2): (Double concave) $[R_1 = -ve, R_2 = +ve]\frac{1}{\text{f}}=(15-1)\Big(\frac{-1}{20}-\frac{1}{30}\Big)\Rightarrow\text{f}=-24\text{cm}$
Case (3): (Concave concave) $[R_1 = -ve, R_2 = -ve]\frac{1}{\text{f}}=(15-1)\Big(\frac{1}{-20}-\frac{1}{-30}\Big)\Rightarrow\text{f}=-120\text{cm}$
Case (4): (Concave convex) $[R_1 = +ve, R_2 = +ve]\frac{1}{\text{f}}=(15-1)\Big(\frac{1}{20}-\frac{1}{30}\Big)\Rightarrow\text{f}=+120\text{cm}$
Case (1): (Double convex) $[R_1 = +ve, R_2 = -ve]\frac{1}{\text{f}}=(15-1)\Big(\frac{1}{20}-\frac{1}{30}\Big)\Rightarrow\text{f}=24\text{cm}$
Case (2): (Double concave) $[R_1 = -ve, R_2 = +ve]\frac{1}{\text{f}}=(15-1)\Big(\frac{-1}{20}-\frac{1}{30}\Big)\Rightarrow\text{f}=-24\text{cm}$
Case (3): (Concave concave) $[R_1 = -ve, R_2 = -ve]\frac{1}{\text{f}}=(15-1)\Big(\frac{1}{-20}-\frac{1}{-30}\Big)\Rightarrow\text{f}=-120\text{cm}$
Case (4): (Concave convex) $[R_1 = +ve, R_2 = +ve]\frac{1}{\text{f}}=(15-1)\Big(\frac{1}{20}-\frac{1}{30}\Big)\Rightarrow\text{f}=+120\text{cm}$
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