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A child with normal near point (25 cm) reads a book with small size print using a magnifying glass: a thin convex lens of focal length 5 cm.
CBSE | Class 12 | NCERT | Physics | Ray Optics and Optical Instruments
A child with normal near point (25 cm) reads a book with small size print using a magnifying glass: a thin convex lens of focal length 5 cm.

(a) What would be the shortest and the longest distance at which the lens should be placed from the page so that the book can be read easily when viewing through the magnifying glass?

(b) What is the max and the mini angular magnification (magnifying power) possible using the above given simple microscope?

Answer ā€“

 (a) Focal length of the given magnifying glass is  f = 5 cm

Distance vision has the least distance which is d = 25 cm

Closest object distance is u

Image distance is v = -d = -25 cm

Using the lens formula, we have ā€“

\frac{1}{f} = \frac{1}{v} ā€“ \frac{1}{u}

\frac{1}{u} = \frac{1}{v} ā€“ \frac{1}{f}

= \frac{1}{-25} ā€“ \frac{1}{5}

= \frac{-6}{25}

Therefore,we have ā€“

 u = \frac{-25}{6}

= -4.167 cm

Hence, 4.167 cm is the closest distance at which the person can read the book.

For the object at the longest distance (uā€™), the image distance (vā€™) is = āˆž

Using the lens formula, we have ā€“

\frac{1}{f} = \frac{1}{v'} ā€“ \frac{1}{u'}

\frac{1}{u'} = \frac{1}{∞} ā€“ \frac{1}{5}

= ā€“ \frac{1}{5}

Therefore, uā€™ is -5 cm

Hence, 5 cm is the farthest distance at which the person can read the book.

(b) Maximum angular magnification is determined by the relation:

\alpha _{max}=\frac{d}{\left | u \right |}

= \frac{25}{\frac{25}{6}}

= 6

Minimum angular magnification is determined by the relation ā€“

= \alpha _{max}=\frac{d}{\left | u' \right |}

= \frac{25}{5}

= 5

i

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