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= ____________ cm

What is the focal length f of this lens in air (index of refraction for air is nair=1)?

Answer 1

Answer:

focal length of the lens when immersed in water is 150cm

focal length of the lens if it is in air is 60cm.

It can be calculated using the equation

1/f = (Refractive index of the glass - Reflective index of the medium)x[ 1/R1 - 1/R2]

Where R1 is the radius of curvature of the surface through which light will enter the lens, and

R2 is the radius of curvature of the surface from which light will exit the lens.

Explanation:

Since n-water = 1.3

focal length of the lens if it is immersed in water is

1/f = (n - 1.3)[1/R2 -1/R1]

1/f = (1.5 - 1.3)[1/10 - 1/15]

1/f = 0.2 x (2/60)

f = 60/0.4 = 150cm

Since n-air = 1

focal length of the lens if it is in air is calculated as:

1/f = (n - 1)[1/R2 -1/R1]

1/f = (1.5 - 1)[1/10 - 1/15]

1/f = 0.5 x (2/60)

f = 60/1= 60cm

Note: The values are measured in centimetre and not converted to metre

Answer 2

Answer: f=150cm in water and f=60cm in air.

Explanation: Focal length is a measurement of how strong light is converged or diverged by a system. To find the variable, it can be used the formula:

$\frac{1}{f}$ = (nglass - ni)($\frac{1}{R1}$ - $\frac{1}{R2}$).

nglass is the index of refraction of the glass;

ni is the index of refraction of the medium you want, water in this case;

R1 is the curvature through which light enters the lens;

R2 is the curvature of the surface which it exits the lens;

Substituting and calculating for water (nwater = 1.3):

$\frac{1}{f}$ = (1.5 - 1.3)($\frac{1}{10}$ - $\frac{1}{15}$)

$\frac{1}{f}$ = 0.2($\frac{1}{30}$)

f = $\frac{30}{0.2}$ = 150

For air (nair = 1):

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

f = $\frac{30}{0.5}$ = 60

In water, the focal length of the lens is f = 150cm.

In air, f = 60cm.