A thin double convex lens has radii of curvat | vedclass

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(B) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For a double convex lens,$R_1 = +40 \;

Vedclass · Accepted Answer

$31$

Vedclass · Answer

(B) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For a double convex lens,$R_1 = +40 \; cm$ and $R_2 = -40 \; cm$.The refractive index $\mu = 1.65$.Substituting these values into the formula:$\frac{1}{f} = (1.65 - 1) \left( \frac{1}{40} - \frac{1}{-40} ight)$$\frac{1}{f} = (0.65) \left( \frac{1}{40} + \frac{1}{40} ight)$$\frac{1}{f} = 0.65 	imes \frac{2}{40} = \frac{0.65}{20} = 0.0325$$f = \frac{1}{0.0325} \approx 30.77 \; cm$.Rounding to the nearest integer,the focal length is approximately $31 \; cm$.

$A$ thin double convex lens has radii of curvature each of magnitude $40 \; cm$ and is made of glass with refractive index $1.65$. Its focal length is nearly....... $cm$.

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