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At What Distance Should the Lens Be Held from the Figure in Exercise 9.29 in Order to View the Squares Distinctly with the Maximum Possible Magnifying Power? - Physics

(a) At what distance should the lens be held from the figure in

Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?

(b) What is the magnification in this case?

(c) Is the magnification equal to the magnifying power in this case?

Explain.

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Solution

(a) The maximum possible magnification is obtained when the image is formed at the near point (d = 25 cm).

Image distance, v = −d = −25 cm

Focal length, f = 10 cm

Object distance = u

According to the lens formula, we have:

`1/f = 1/v - 1/u`

`1/u = 1/v - 1/f`

`= 1/(-25) - 1/10 = (-2-5)/50 = - 7/50`

`:. u  = 50/7 = -7.14 cm`

Hence, to view the squares distinctly, the lens should be kept 7.14 cm away from them.

(b) Magnification = `|v/u| = 25/50 = 3.5`

(c) Magnifying power  = `d/u = 25/(50/7) = 3.5` 

Since the image is formed at the near point (25 cm), the magnifying power is equal to the magnitude of magnification.

  Is there an error in this question or solution?
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APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 9 Ray Optics and Optical Instruments
Q 30 | Page 348
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