Question

A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvature with magnitudes of |R1|=10cm and |R2|=15cm. The lens is made of glass with index of refraction nglass=1.5. We will employ the convention that R1 refers to the radius of curvature of the surface through which light will enter the lens, and R2 refers to the radius of curvature of the surface from which light will exit the lens.Part AIs this lens converging or diverging?Part BWhat is the focal length f of this lens in air (index of refraction for air is nair=1)?Express your answer in centimeters to two significant figures or as a fraction.

Answer 1

Answer:

12 cm

Explanation:

We shall use Lens makers formula here which is as follows

$\frac{1}{F} =(\mu-1) (\frac{1}{R_1} -\frac{1}{R_2})$

Put μ = 1.5 , R₁ = 10 cm ,R₂ = -  15 cm ( according to sign convention )

$\frac{1}{F} =(1.5-1) (\frac{1}{10} -\frac{1}{-15})$

= .5 x ( 15 + 10 ) / 15 x 10

= $\frac{25}{2\times10\times15}$

F = 12 cm