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.C) What is the focal length of the lens if it is immersed in water (nwater = 1.3)f= ____________ cmWhat is the focal length f of this lens in air (index of refraction for air is nair=1)?

Answer 1

Answer:

for air focus is 12 cm

when immersed in water focus is 39 c.m

Explanation:

we know lens maker formula $\frac{1}{f}=\frac{n2-n1}{n1}$

so for air n=1 and n2=1.5

now putting the value in lens maker formula $[tex][tex]\frac{1}{f}=\frac{1.5-1}{1}(\frac{1}{.1}  - \frac{1}{-.15} )$now $\frac{1}{f} =\frac{25}{3} \\\\f=\frac{3}{25} \\\\f=0.12 m\\\\f=12 c.m$[/tex][/tex]

when lens is immersed in water then n1=1.3 and n2=1.5

putting the value in lens maker formula

$[tex]\frac{1}{f}=\frac{1.5-1.3}{1.3}(\frac{1}{.1}  - \frac{1}{-.15} )$now $\frac{1}{f} =\frac{100}{39} \\\\f=\frac{39}{100} \\\\f=0.39 m\\\\f=39 c.m$[/tex]