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 (n_water = 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

Concepts and reason

The concept used in this question is lens maker formula.

Firstly, write the expression of the lens maker formula and use it to find the focal length of the lens when it is immersed in water.

Finally, use the formula to find the focal length of the lens when it is placed in air.

Fundamentals

According to the thin lens formula, the focal length of the lens when it is placed in a medium can be determined by the following expression:

$\frac{1}{f} = \left( {\frac{{{n_{\rm{L}}}}}{{{n_{\rm{m}}}}} - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)$

Here, f is the focal length of the lens, ${n_{\rm{L}}}$ is the refractive index of the lens, ${n_{\rm{m}}}$ is the refractive index of the medium, ${R_1}$ , and ${R_2}$ are the radius of curvature of the lens through which the

(a)

The lens maker formula is as follows:

$\frac{1}{f} = \left( {\frac{{{n_{\rm{L}}}}}{{{n_{\rm{m}}}}} - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)$

The lens is immersed in water so that the medium is water. The radius of curvature of the two curved surfaces of the lens are as follows:

$\begin{array}{c}\\{R_1} = 10{\rm{ cm}}\\\\{R_2} = - 15{\rm{ cm}}\\\end{array}$

Substitute 1.5 for ${n_{\rm{L}}}$ , 1.3 for ${n_{\rm{m}}}$ , 10 cm for ${R_1}$ , and -15 cm for ${R_2}$ in the above expression:

$\begin{array}{c}\\\frac{1}{f} = \left( {\frac{{1.5}}{{1.3}} - 1} \right)\left( {\frac{1}{{10{\rm{ cm}}}} - \frac{1}{{ - 15{\rm{ cm}}}}} \right)\\\\ = \left( {0.153846} \right)\left( {0.1 + 0.0666666} \right){\rm{ c}}{{\rm{m}}^{ - 1}}\\\\ = 0.025641{\rm{ c}}{{\rm{m}}^{ - 1}}\\\\f = 39{\rm{ cm}}\\\end{array}$

(b)

The lens maker formula is as follows:

$\frac{1}{f} = \left( {\frac{{{n_{\rm{L}}}}}{{{n_{\rm{m}}}}} - 1} \right)\left( {\frac{1}{{{R_1}}} - \frac{1}{{{R_2}}}} \right)$

The lens is placed in air so that the medium is air. The radius of curvature of the two curved surfaces of the lens are as follows:

$\begin{array}{c}\\{R_1} = 10{\rm{ cm}}\\\\{R_2} = - 15{\rm{ cm}}\\\end{array}$

Substitute 1.5 for ${n_{\rm{L}}}$ , 1.0 for ${n_{\rm{m}}}$ , 10 cm for ${R_1}$ , and -15 cm for ${R_2}$ in the above expression:

$\begin{array}{c}\\\frac{1}{f} = \left( {\frac{{1.5}}{{1.0}} - 1} \right)\left( {\frac{1}{{10{\rm{ cm}}}} - \frac{1}{{ - 15{\rm{ cm}}}}} \right)\\\\ = \left( {0.5} \right)\left( {0.1 + 0.0666666} \right){\rm{ c}}{{\rm{m}}^{ - 1}}\\\\ = 0.083333{\rm{ c}}{{\rm{m}}^{ - 1}}\\\\f = 12{\rm{ cm}}\\\end{array}$

Ans: Part a

The focal length is 39 cm.

Part b

The focal length is 12 cm.