Question

The distance between an object and a screen is 98cm. A convex lens placed between object and screen so as it get an image on the screen magnified 12 times. Calculate the focal length and power of lens.

Answer 1

Answer on Question #73641 Physics / Optics

The distance between an object and a screen is $98~\mathrm{cm}$. A convex lens placed between object and screen so as it get an image on the screen magnified 12 times. Calculate the focal length and power of lens.

Solution:

For a thin lens the focal length $f$ is given by equation

$$\frac {1}{f} = \frac {1}{u} + \frac {1}{v}$$

where $u$ is the distance from the object to the lens, $v$ is the distance from the lens to the image (screen).

In our case

$$\left\{ \begin{array}{l} u + v = 0.98~\mathrm{m} \\ \frac {v}{u} = 12 \end{array} \right.$$

We get

$$u = 0.075~\mathrm{m}, \quad v = 0.905~\mathrm{m}$$

Thus

$$\frac {1}{f} = \frac {1}{0.075} + \frac {1}{0.905}$$ $$f = 0.07~\mathrm{m}$$

The power of lens

$$P = \frac {1}{f} = \frac {1}{0.07} = 14.4~\mathrm{m}^{-1}$$

Answers:

$$f = 0.07~\mathrm{m}$$ $$P = 14.4~\mathrm{m}^{-1}$$

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