Ncert Exercises & Solutions

9.34:

A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when

(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?

(b) the final image is formed at the least distance of distinct vision (25 cm)?

(a)
Hint: $m=\frac{f_{0}}{f_{e}}$

Step: Find the magnifying power in normal adjustment.
The focal length of the objective lens, $f_{0}$= 140 cm
The focal length of the eyepiece, f $_{}$ _e = 5 cm
Least distance of distinct vision, d = 25 cm
For the normal adjustment, the magnifying power:

$m = \frac{f_{0}}{f_{e}} = \frac{140}{5} = 28$

(b)
Hint: $m=\frac{f_{0}}{f_{e}}(1+\frac{f_{e}}{d})$

Step: Find the magnifying power for the final image at the least distance of distinct vision.

For the final image at d, the magnifying power:

$\frac{f_{0}}{f_{e}} [1 + \frac{f_{e}}{d}] = \frac{140}{5} [1 + \frac{5}{25}] = 28 [1 + 0.2] = 28 \times 1.2 = 33.6$

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