Question

A convex meniscus lens made of glass ($n_l = 1.51$) has $|R_1| = 9.0$ cm. and $|R_2| = 11.0$ cm. What is the focal distance of this lens in air? Lens equation is: $\frac{1}{s_o} + \frac{1}{s_i} = \frac{n_l - n_m}{n_m} \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

          A convex meniscus lens made of glass ($n_l = 1.51$) has $|R_1| = 9.0$ cm. and $|R_2| = 11.0$ cm.

What is the focal distance of this lens in air?
Lens equation is:
$\frac{1}{s_o} + \frac{1}{s_i} = \frac{n_l - n_m}{n_m} \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

A convex meniscus lens made of glass (nl = 1.51) has |R1| = 9.0 cm. and |R2| = 11.0 cm.

What is the focal distance of this lens in air?
Lens equation is:
(1)/(so) + (1)/(si) = (nl - nm)/(nm)((1)/(R1) - (1)/(R2))

Added by Patricia V.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

06/19/2023

Step 1

5 (refractive index of glass) R1 = 9.0 cm (radius of curvature of the first surface) R2 = 11.0 cm (radius of curvature of the second surface) Show more…

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A convex meniscus lens made of glass n=1.5 has R1=9.0 cm and R2=11.0 cm. What is the focal distance of this lens in air? The lens equation is: 1/f = (n-1) * (1/R1 - 1/R2)