A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.

(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?

(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?

Question

Verified by Toppr

Answer 1

(a)
$f = 25 c m$
least distance of distance vision, $d = 25 c m$
closed object distance is $u$ .
Image distance $v = - d = - 25 c m$ .
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow u = - 4.167 \approx - 4.2$
Hence closest distance to resd book is 4.2 cm

for the farthest distance, $v = \infty$
$\frac{1}{v} - \frac{1}{u _{1}} = \frac{1}{f} \Rightarrow u_{1} = - 5$

Hence the farthest distance at which person can read the book is $5 c m$ .

(b)

maximum angular magnification,

$α = d / ∣ u ∣ = 25 / ∣ - 4.167 ∣ = 6$

minimum angular magnification,

$α = d / ∣ u_{1} ∣ = 25 / ∣ 5 ∣ = 5$

Answer 2

(a)

f = 25 c m

least distance of distance vision,

d = 25 c m

closed object distance is

u

.

Image distance

v = - d = - 25 c m

.

\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow u = - 4.167 \approx - 4.2

Hence closest distance to resd book is 4.2 cm

for the farthest distance,

v = \infty

\frac{1}{v} - \frac{1}{u _{1}} = \frac{1}{f} \Rightarrow u_{1} = - 5

Hence the farthest distance at which person can read the book is

5 c m

.

(b)

maximum angular magnification,

$α = d / ∣ u ∣ = 25 / ∣ - 4.167 ∣ = 6$

minimum angular magnification,

$α = d / ∣ u_{1} ∣ = 25 / ∣ 5 ∣ = 5$