A plane convex lens has a refractive index of $1.6$ and a radius of curvature $60cm$ . What is the focal length of the lens?
Answer
546.6k+ views
Hint:In order of this question, first we will rewrite the given facts given in the question, and then we will apply the formula that relates the focal length, refractive index and the radius of curvature, to find the focal length of the given lens.
Complete step-by-step solution:
Radius of curvature of a plane convex lens, ${R_1} = 60cm$
The radius of curvature is positive if the vertex is to the left of the centre of curvature. The radius of curvature is negative if the vertex is to the right of the centre of curvature. When looking at a biconvex lens from the side, the left surface radius of curvature is positive, while the right is negative.
$\because {R_2} = \infty $
Refractive index of a plane convex lens, $\mu = 1.6$
The refractive index of the lens material in relation to the surrounding material determines the focal length of a narrow lens. Assume that one side of the lens has an absolute refractive index of ${n_1}$ and the other side has an absolute refractive index of ${n_3}$ .
Now, we will apply the relationship between focal length, refractive index and the radius of curvature:
$
\therefore \dfrac{1}{f} = (\mu - 1)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}) \\
\Rightarrow \dfrac{1}{f} = (1.6 - 1)(\dfrac{1}{{60}}) \\
\Rightarrow f = 100cm \\
$
Hence, the focal length of the lens is $100cm$ .
Note:The radius of curvature is negative if the vertex is to the right of the centre of curvature. When looking at a biconvex lens from the side, the left surface radius of curvature is positive, while the right is negative.
Complete step-by-step solution:
Radius of curvature of a plane convex lens, ${R_1} = 60cm$
The radius of curvature is positive if the vertex is to the left of the centre of curvature. The radius of curvature is negative if the vertex is to the right of the centre of curvature. When looking at a biconvex lens from the side, the left surface radius of curvature is positive, while the right is negative.
$\because {R_2} = \infty $
Refractive index of a plane convex lens, $\mu = 1.6$
The refractive index of the lens material in relation to the surrounding material determines the focal length of a narrow lens. Assume that one side of the lens has an absolute refractive index of ${n_1}$ and the other side has an absolute refractive index of ${n_3}$ .
Now, we will apply the relationship between focal length, refractive index and the radius of curvature:
$
\therefore \dfrac{1}{f} = (\mu - 1)(\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}) \\
\Rightarrow \dfrac{1}{f} = (1.6 - 1)(\dfrac{1}{{60}}) \\
\Rightarrow f = 100cm \\
$
Hence, the focal length of the lens is $100cm$ .
Note:The radius of curvature is negative if the vertex is to the right of the centre of curvature. When looking at a biconvex lens from the side, the left surface radius of curvature is positive, while the right is negative.
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