# What should be the distance between the object and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. - Physics

Numerical

What should be the distance between the object and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier?

#### Solution

Area of the virtual image of each square, A = 6.25 mm2

Area of each square, A0 = 1 mm2

Hence, the linear magnification of the object can be calculated as:

m = sqrt("A"/"A"_0)

= sqrt((6.25)/1)

= 2.5

But m = ("Image distance (v)")/"Object distnace (u)"

∴ v = mu = 2.5 u ............(1)

Focal length of the magnifying glass, f = 10 cm

According to the lens formula, we have the relation:

1/"f" = 1/"v" - 1/"u"

1/10 = 1/(2.5"u") - 1/"u" = 1/"u" (1/2.5 - 1/1) = 1/"u" ((1-2.5)/2.5)

∴ u = - (1.5  xx 10)/2.5 = −6 cm

And v = 2.5 u

= 2.5 × 6 = −15 cm

The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.

Concept: Optical Instruments - The Eye
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 9 Ray Optics and Optical Instruments
Exercise | Q 9.25 | Page 347
NCERT Class 12 Physics Textbook
Chapter 9 Ray Optics and Optical Instruments
Exercise | Q 31 | Page 348
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