
What is the focal length of a double convex lens for which the radius of curvature of each surface is 60 cm (n=1.5)?
A. 50 cm
B. 60 cm
C. 90 cm
D. 30 cm
Answer
571.2k+ views
Hint: Generally, if nothing is given about the lens, we take it to be thin and having focal length independent of the refractive index. In that case, we take focal length as half of the radius of curvature. But practically, the lens has thickness and the parameters depend upon the refractive index also.
Formula used:
$\dfrac 1f = \left( \dfrac{n_2}{n_1} -1\right) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$
Complete step by step answer:
For computing the focal length of any lens, we practically use the formula called lens makers formula. This formula is given by $\dfrac 1f = \left( \dfrac{n_2}{n_1} -1\right) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$
Here, $R_1\ and\ R_2$are radii of curvature of two faces of the lens. $n_2$ is the refractive index of material and $n_1$ is the refractive index of the medium in which the lens is kept, as shown in the figure.
Given, $n_2 = 1.5,\ n_1 = 1\ (by\ default)$
Also given that the radii of curvature for both sides is the same. But, for the left side, we will take radius to be positive whereas for the right side, we will take radius to be negative (as per the sign conventions).
Hence $R_1 = -60cm\ and\ R_2 = +60cm$
Putting the values in the equation: $\dfrac1f = \left( \dfrac{n_2}{n_1} -1\right) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$, we get:
$\dfrac1f = \left( \dfrac{1.5}{1} -1\right) \left( \dfrac{1}{-60} - \dfrac{1}{60}\right)$
$\implies \dfrac1f = (0.5)\left( -\dfrac{2}{60} \right) = \dfrac{-1}{60}$
Or $f = -60 cm$
Hence, the focal length of the double convex lens is -60 cm.
Note:
Students should note that in case of a thin lens, we directly take ‘f’ as half of ‘R’. But in this question, we have the value of ‘R’ and ‘f’ is the same. One must not confuse between a thin and practical lens. Whenever the refractive index of the material is given and the lens is not thin, we have to calculate the focal length of the lens using the lens makers formula.
Formula used:
$\dfrac 1f = \left( \dfrac{n_2}{n_1} -1\right) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$
Complete step by step answer:
For computing the focal length of any lens, we practically use the formula called lens makers formula. This formula is given by $\dfrac 1f = \left( \dfrac{n_2}{n_1} -1\right) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$
Here, $R_1\ and\ R_2$are radii of curvature of two faces of the lens. $n_2$ is the refractive index of material and $n_1$ is the refractive index of the medium in which the lens is kept, as shown in the figure.
Given, $n_2 = 1.5,\ n_1 = 1\ (by\ default)$
Also given that the radii of curvature for both sides is the same. But, for the left side, we will take radius to be positive whereas for the right side, we will take radius to be negative (as per the sign conventions).
Hence $R_1 = -60cm\ and\ R_2 = +60cm$
Putting the values in the equation: $\dfrac1f = \left( \dfrac{n_2}{n_1} -1\right) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$, we get:
$\dfrac1f = \left( \dfrac{1.5}{1} -1\right) \left( \dfrac{1}{-60} - \dfrac{1}{60}\right)$
$\implies \dfrac1f = (0.5)\left( -\dfrac{2}{60} \right) = \dfrac{-1}{60}$
Or $f = -60 cm$
Hence, the focal length of the double convex lens is -60 cm.
Note:
Students should note that in case of a thin lens, we directly take ‘f’ as half of ‘R’. But in this question, we have the value of ‘R’ and ‘f’ is the same. One must not confuse between a thin and practical lens. Whenever the refractive index of the material is given and the lens is not thin, we have to calculate the focal length of the lens using the lens makers formula.
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