Question

A planoconvex lens is made of glass of refractive index 1.5. The radius of curvature of its spherical surface is 40 cm. At what distance from the lens will a parallel beam of light will get focussed? Draw a neat, clear labelled ray diagram to show image formation from the lens. What will be the nature and power of a combination of two such lenses kept with their (i) plane surfaces in contact (ii) spherical surfaces in contact.

Answer 1

Answer on Question #57184, Physics / Optics

A planoconvex lens is made of glass of refractive index 1.5. The radius of curvature of its spherical surface is $40~\mathrm{cm}$. At what distance from the lens will a parallel beam of light will get focussed? Draw a neat, clear labelled ray diagram to show image formation from the lens. What will be the nature and power of a combination of two such lenses kept with their (i) plane surfaces in contact (ii) spherical surfaces in contact.

Solution:

For a lens in air, the power $1/f$ is then given by

$$\frac{1}{f} = (n - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$$

The radii of curvature here are measured according to the Cartesian sign convention.

In our case,

$$R_1 = 40~\mathrm{cm}, \quad R_2 = \infty$$

Thus,

$$f = \frac{R_1}{n - 1} = \frac{40}{1.5 - 1} = 80~\mathrm{cm}$$

A parallel beam of light will be focused at $80~\mathrm{cm}$ from the lens.

(i)

For a double convex lens the radius $R_1$ is positive since it is measured from the front surface and extends right to the center of curvature. The radius $R_2$ is negative since it extends left from the second surface.

$$\frac{1}{f} = (n - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right] = (1.5 - 1) \left[ \frac{1}{0.40} + \frac{1}{0.40} \right] = 2.5~\mathrm{D}$$

$1/f > 0$ and this lens will be converging.

(ii)

When spherical surfaces in contact the combined focal length $f$ of the lenses is given by

$$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{0.8} + \frac{1}{0.8} = 2.5~\mathrm{D}$$

$1/f > 0$ and this lens will be converging.

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