**1.** 1. A person looks at different trees in an open space with
the following details. Arrange the trees in decreasing
order of their apparent sizes. $\\$
Tree

The visual angles made by the trees with the eyes can be calculated be below. $\\$ $\theta = \frac{Height \ of \ the \ tree}{Distance \ from \ the \ eye } = \frac{AB}{OB} \Rightarrow \theta_A = \frac{2}{50} = 0.04$ $\\$ simillerly, $\theta_B= \frac{2.5}{80} = 0.03125$ $\\$ $\theta_C= \frac{1.8}{70} = 0.02571$ $\\$ $\theta_D= \frac{2.8}{100}=0.028$ $\\$ Since, $\theta_A > \theta_B > \theta_D > \theta_C $, the arrangement in decreasing order is given by A, B, D and C.

**2.** An object is to be seen through a simple microscope of
focal length $12$ cm. Where should the object be placed
so as to produce maximum angular magnification ? The
least distance for clear vision is $25$ cm.

For the given simple microscope, $\\$ f = 12 cm and D = 25 cm $\\$ For maximum angular magnification, the image should be produced at least distance of clear vision. $\\$ So, $v= -D= -25 \ cm$ $\\$ Now, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ $\\$ $\Rightarrow \frac{1}{u}=\frac{1}{v}-\frac{1}{f} = \frac{1}{-25}-\frac{1}{12}= -\frac{37}{300}$ $\\$ $\Rightarrow u = -8.1 \ cm$ $\\$ So, the object should be placed 8.1 cm away from the lens.

**3.** A simple microscope has a magnifying power of $3.0$ when
the image is formed at the near point $(25 cm)$ of a normal
eye. (a) What is its focal length ? (b) What will be its
magnifying power if the image is formed at infinity ?

The simple microscope has, m = 3, when image is formed at D = 25 cm $\\$ a) $m=1+\frac{D}{f} \Rightarrow 3 = 1+\frac{25}{f}$ $\\$ $\Rightarrow f = 25/2 = 12.5 \ cm$ $\\$ b) when the image is formed at infinity (normal adjustment ) $\\$ Magnifying power = $\frac{D}{f} = \frac{25}{12.5} = 2.0$ $\\$

**4.** A child has near point at $10$ cm. What is the maximum
angular magnification the child can have with a convex
lens of focal length $10$ cm ?

The child has D = 10 cm and f = 10 cm $\\$ The maximum angular magnification is obtained when the image is formed at near point. $\\$ $m = 1+\frac{D}{f} = 1 + \frac{10}{10} = 1+1=2$

**5.** A simple microscope is rated $5$ X for a normal relaxed
eye. What will be its magnifying power for a relaxed
farsighted eye whose near point is $40$ cm ?

The simple microscope has magnification of 5 for normal relaxed eye (D = 25 cm). $\\$ Because, the eye is relaxed the image is formed at infinity (normal adjustment) $\\$ So, $m=5 = \frac{D}{f} = \frac{25}{f} \Rightarrow f=5 cm$ $\\$ For the relaxed farsighted eye, D = 40 cm $\\$ So, $m = \frac{D}{f} = \frac{40}{5} = 8$ $\\$ So, its magnifying power is 8X

**6.** Find the maximum magnifying power of a compound
Microscope having a $25$ diopter lens as the objective, a
$5$ diopter lens as the eyepiece and the separation $30$ cm
between the two lenses. The least distance for clear
vision is $25$ cm.

For the given compound microscope $\\$ $f_0 = \frac{1}{25 \ diopter} = 0.04 \ m = 4 \ cm, f_e = \frac{1}{5 \ diopter} = 0.2 \ m = 20 \ cm $ $\\$ $D = 25 cm,$ separation between objective and eyepiece = 30 cm $\\$ The magnifying power is maximum when the image is form by $\\$ the eye piece at least distance of clear vision i.e. D = 25 cm $\\$ for the eye piece, $v_e = -25 \ cm, \ f= 20 \ cm$ $\\$ For lens formula, $\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}$ $\\$ $\Rightarrow \frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e} \Rightarrow \frac{1}{-25}-\frac{1}{20} \Rightarrow u_e = 11.11 cm$ $\\$ So, for the objective lens, the image distance should be $\\$ $v_D = 30-(11.11) = 18.89 \ cm$ $\\$ Now , for the objective lens, $\\$ $v_0 = +18.89 \ cm \ $ (Because real image is produced) $\\$ $f_D = 4 \ cm$ $\\$ So, $\frac{1}{u_0} = \frac{1}{v_0}-\frac{1}{f_0} \Rightarrow \frac{1}{18.89}-\frac{1}{4} = 0.053-0.25=-0.197$ $\\$ $\Rightarrow u_0=-5.07 cm$ $\\$ So, the maximum magnificent power is given by $\\$ $m = -\frac{u_0}{v_0} \Bigg[ 1+\frac{D}{f_e} \Bigg] = -\frac{18.89}{-5.07} \Bigg[ 1+\frac{25}{20} \Bigg] $ $\\$ $= 3.7225 \times 2.25 = 8.376$

**7.** The separation between the objective and the eyepiece
of a compound microscope can be adjusted between
$9.8$ cm to $11.8$ cm. If the focal lengths of the objective
and the eyepiece are $1.0$ cm and $6$ cm respectively, find
the range of the magnifying power if the image is always
needed at $24$ cm from the eye

For the given compound microscope $\\$ $f_o$ = 1 cm, $f_e = $ 6 cm, D = 24 cm $\\$ For the eye piece, $v_e = -24$ cm, $f_e = $ 6 cm $\\$ Now, $\frac{1}{v_e}-\frac{1}{u_e} = \frac{1}{f_e}$ $\\$ $\Rightarrow \frac{1}{u_e} = \frac{1}{v_e} - \frac{1}{f_e} \Rightarrow - \bigg[ \frac{1}{24} + \frac{1}{6} \bigg] = -\frac{5}{24}$ $\\$ $\Rightarrow u_e = -4.8 \ cm$ $\\$ a) When the separation between objective and eye piece is 9.8 cm, the image distance for the objective lens must be (9.8)-(4.8) = 5.0 cm $\\$ Now, $\frac{1}{v_0}-\frac{1}{u_0} = \frac{1}{f_0}$ $\\$ $\Rightarrow \frac{1}{u_0} = \frac{1}{v_0}-\frac{1}{f_0} = \frac{1}{5}-\frac{1}{1} = -\frac{4}{5}$ $\\$ $\Rightarrow = u_0 = -\frac{5}{4} = -1.25 \ cm$ $\\$ So, the magnifying power is given by, $\\$ $m = \frac{v_0}{u_0} \Bigg[ 1+ \frac{D}{f} \Bigg] = \frac{-5}{-1.25} \Bigg[ 1+\frac{24}{6} \Bigg] = 4 \times 5 = 20 $ $\\$ b) When the separation is 11.8 cm, $\\$ $v_0 = 11.8 - 4.8 = 7.0 cm, \qquad f_0 = 1 \ cm$ $\\$ $\Rightarrow \frac{1}{u_0} = \frac{1}{v_0}-\frac{1}{f_0} = \frac{1}{7} -\frac{1}{1} = -\frac{6}{7}$ $\\$ So, $m = -\frac{v_0}{u_0} \Bigg[ 1+ \frac{D}{f} \Bigg] = \frac{-7}{- (\frac{7}{6})} \Bigg[ 1+ \frac{24}{6} \Bigg] = 6 \times 5 = 30$ $\\$ So, the range of magnifying power will be 20 to 30.

**8.** An eye can distinguish between two points of an object
if they are separated by more than $0.22$ mm when the
object is placed at $25$ cm from the eye. The object is now
seen by a compound microscope having a $20$ $D$ objective
and $10$ $D$ eyepiece separated by a distance of $20$ $cm.$ The
final image is formed at $25$ $cm$ from the eye. What is
the minimum separation between two points of the
object which can now be distinguished ?

For the given compound microscope. $\\$ $f_0=\frac{1}{20D}=$ 0.05 m = 5 cm, $\qquad f_e= \frac{1}{10D}=$ 0.1 m = 10 cm. $\\$ D = 25 cm, separation between objective $\&$ eyepiece= 20 cm For the minimum separation between two points which can be distinguished by eye using the microscope, the magnifying power should be maximum. $\\$ For the eyepiece, $v_0 = –25 \ cm, f_e = 10 \ cm$ $\\$ So, $\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e} = \frac{1}{-25}-\frac{1}{10} = -\bigg[ \frac{2+5}{50} \bigg ] \Rightarrow u_e = -\frac{50}{7}cm$ $\\$ So, the image distance for the objective lens should be, $\\$ $v_0 = 20 - \frac{50}{7} = \frac{90}{7} \ cm$ $\\$ Now, for the objective lens, $\\$ $\frac{1}{u_0} = \frac{1}{v_0}-\frac{1}{f_0} = \frac{7}{90}-\frac{1}{5} = -\frac{11}{90}$ $\\$ $\Rightarrow u_0 = -\frac{90}{11}cm$ $\\$ So, the maximum magnifying power is given by, $\\$ $m = \frac{-v_0}{u_0} \Bigg[ 1+\frac{D}{f_e} \Bigg ]$ $\\$ $= \frac{ \bigg( \frac{90}{7} \bigg) }{\bigg( -\frac{90}{11} \bigg)} \Bigg[ 1+\frac{25}{10} \Bigg]$ $\\$ $= \frac{11}{7} \times 3.5 = 5.5$ $\\$ Thus, minimum separation eye can distinguish $= \frac{0.22}{5.5}mm = 0.04 \ mm$ $\\$

**9.** A compound microscope has a magnifying power of $100$
when the image is formed at infinity. The objective has
a focal length of $0.5$ cm and the tube length is $6.5$ $cm.$
Find the focal length of the eyepiece.

For the give compound microscope, $\\$ $f_0$ = 0.5cm, tube length = 6.5 cm $\\$ magnifying power = 100 (normal adjustment) $\\$ Since, the image is formed at infinity, the real image produced by the $\\$ objective lens should lie on the $\\$ focus of the eye piece. $\\$ So, $v_0 + f_e =$ 6.5 cm $\qquad$ ...(1) $\\$ Again, magnifying power $= \frac{v_0}{u_0} \times \frac{D}{f_e} \qquad$ [for normal adjustment] $\\$ $\Rightarrow m =- \Bigg[ 1- \frac{v_0}{f_0} \Bigg] \frac{D}{f_e} \qquad \quad \Bigg[ \because \frac{v_0}{u_0} = 1-\frac{v_0}{f_0} \Bigg]$ $\\$ $\Rightarrow 100 = - \Bigg[ 1- \frac{v_0}{0.5} \Bigg] \times \frac{25}{f_e} \qquad \quad$ [Taking D = 25 cm] $\\$ $\Rightarrow 100 \ f_e = -(1-2 v_0) \times 25$ $\\$ $\Rightarrow 2v_0 - 4f_e = 1 \qquad \qquad$ ...(2) $\\$ Solving equation (1) and (2) we can get, $\\$ $V_0$ = 4.5 cm and $f_e$ = 2 cm $\\$ So, the focal length of the eye piece is 2cm.

**10.** A compound microscope consists of an objective of focal
length $1$ cm and an eyepiece of focal length $5$ cm. An
object is placed at a distance of $0.5$ cm from the objective.
What should be the separation between the lenses so
that the microscope projects an inverted real image of
the object on a screen $30$ cm behind the eyepiece ?

Given that, $\\$ $f_o$ = = 1 cm, $f_e$ = 5 cm, $\qquad$ $u_0$ = 0.5 cm, $\qquad$ $v_e$ = 30 cm $\\$ For the objective lens, $u_0$ = – 0.5 cm, $f_0$ = 1 cm. $\\$ From lens formula, $\\$ $\frac{1}{v_0} - \frac{1}{u_0} = \frac{1}{f_0} \qquad \Rightarrow \frac{1}{v_0} = \frac{1}{u_0} + \frac{1}{f_0} = \frac{1}{-0.5} + \frac{1}{1} = -1$ $\\$ $\Rightarrow v_0 = -1$ cm $\\$ So, a virtual image is formed by the objective on the same side as that of the object at a distance of 1 cm from the objective lens. This image acts as a virtual object for the eyepiece. $\\$ For the eyepiece, $\\$ $\frac{1}{v_0} - \frac{1}{u_0} = \frac{1}{f_0} \qquad \Rightarrow \frac{1}{u_0} = \frac{1}{v_0}-\frac{1}{f_0} = \frac{1}{30} - \frac{1}{5} = \frac{-5}{30} = \frac{-1}{6}$ $\\$ $\Rightarrow u_0 = -6$ cm $\\$ So, as shown in figure,$\\$ Separation between the lenses = $u_0 – v_0$ = 6 – 1 = 5 cm

**11.** An optical instrument used for angular magnification
has a $25$ D objective and a $20$ D eyepiece. The tube
length is $25$ cm when the eye is least strained. $\\$
(a) Whether it is a microscope or a telescope ? (b) What
is the angular magnification produced ?

The optical instrument has $\\$ $f_0 = \frac{1}{25D} = $ 0.04 m = 4 cm $\\$ $f_e = \frac{1}{20D} = $ 0.05 m = 5 cm $\\$ tube length = 25 cm (normal adjustment) $\\$ (a) The instrument must be a microscope as $f_0 < f_e$ $\\$ (b) Since the final image is formed at infinity, the image produced by the objective should lie on the focal plane of the eye piece. $\\$ So, image distance for objective = $v_0$ = 25 – 5 = 20 cm $\\$ Now, using lens formula. $\\$ $\frac{1}{v_0} - \frac{1}{u_0} = \frac{1}{f_0} \qquad \Rightarrow \frac{1}{u_0} = \frac{1}{v_0} - \frac{1}{f_0} = \frac{1}{20} - \frac{1}{4} = \frac{-4}{20} = \frac{-1}{5}$ $\\$ $\Rightarrow u_0 = -5 \ cm$ $\\$ So, angular magnification = m $= \frac{v_0}{u_0} \times \frac{D}{f_e} \quad $ [Taking D = 25 cm] $\\$ $= -\frac{20}{-5} \times \frac{25}{5} = 20$

**12.** An astronomical telescope is to be designed to have a
magnifying power of $50$ in normal adjustment. If the
length of the tube is $102$ cm, find the powers of the
objective and the eyepiece.

For the astronomical telescope in normal adjustment.$\\$ Magnifying power = m = 50, length of the tube = L = 102 cm $\\$ Let $f_0$ and $f_e$ be the focal length of objective and eye piece respectively. $\\$ $m = \frac{f_0}{f_e} = 50 \Rightarrow f_0 = 50 \ f_e \qquad$ ...(1) $\\$ and, L = $f_0$ + $f_e$ = 102 cm $\qquad$ ...(2) $\\$ Putting the value of $f_0$ from equation (1) in (2), we get, $\\$ $f_0 + f_e =102 \Rightarrow 51 f_e = 102 \Rightarrow f_e = 2 \ cm = 0.02 \ m$ $\\$ So, $f_0$ = 100 cm = 1 m $\\$ $\because$ Power of the objective lens = $\frac{1}{f_0} = 1D$ $\\$ And Power of the eye piece lens $= \frac{1}{f_e} = \frac{1}{0.02} = 50 D$

**13.** The eyepiece of an astronomical telescope has a focal
length of $10$ cm. The telescope is focused for normal
vision of distant objects when the tube length is $1.0$ m.
Find the focal length of the objective and the magnifying
power of the telescope

For the given astronomical telescope in normal adjustment, $\\$ $F_e$ = 10 cm, $\qquad$ L = 1 m = 100cm $\\$ S0, $f_0$ = L – $f_e$ = 100 – 10 = 90 cm $\\$ and, magnifying power $= \frac{f_0}{f_e}=\frac{90}{10} = 9$ $\\$

**14.** A Galilean telescope is $27$ cm long when focused to form
an image at infinity. If the objective has a focal length
of $30$ cm, what is the focal length of the eyepiece ?

For the given Galilean telescope, (When the image is formed at infinity) $\\$ $f_0$ = 30 cm, $\qquad$ L = 27 cm $\\$ Since L $= f_0 – \mid f_e \mid $ $\\$ [Since, concave eyepiece lens is used in Galilean Telescope] $\\$ $\Rightarrow f_e = f_0 - L=$ 30 – 27 = 3 cm $\\$

**15.** A farsighted person cannot see objects placed closer to
$50$ cm. Find the power of the lens needed to see the
objects at $20$ cm.

For the far sighted person, $\\$ u = – 20 cm, $\qquad$ v = – 50 cm $\\$ from lens formula $\frac{1}{v}-\frac{1}{u} = \frac{1}{f}$ $\\$ $\frac{1}{f} = \frac{1}{-50} - \frac{1}{-20} = \frac{1}{20}-\frac{1}{50} = \frac{3}{100}$ $\\$ $\Rightarrow f = \frac{100}{3} \ cm = \frac{1}{3} \ m$ $\\$ So, power of the lens $= \frac{1}{f} = 3 $ Diopter $\\$

**16.** A nearsighted person cannot clearly see beyond $200$ cm.
Find the power of the lens needed to see objects at large
distances.

For the near sighted person, $\\$ u = $\infty$ and v = – 200 cm = – 2m $\\$ So, $\frac{1}{f} = \frac{1}{v}- \frac{1}{u} = \frac{1}{-2} - \frac{1}{\infty} = -\frac{1}{2} = -0.5 $ $\\$ So, power of the lens is –0.5D

**17.** A person wears glasses of power $- 2.5$ D. Is the person
farsighted or nearsighted ? What is the far point of the
person without the glasses ?

The person wears glasses of power –2.5D $\\$ So, the person must be near sighted. $\\$ $u = \infty \qquad$ v = far point, $\qquad$ $f = \frac{1}{-2.5} = -0.4m = – 40 \ cm$ $\\$ Now, $\frac{1}{v}- \frac{1}{u} = \frac{1}{f}$ $\\$ $\Rightarrow \frac{1}{v} = \frac{1}{u} + \frac{1}{f} = 0+\frac{1}{-40} \qquad \Rightarrow v = – 40 \ cm$ $\\$ So, the far point of the person is 40 cm

**18.** A professor reads a greeting card received on his 50th
birthday with $+$ $2.5$ D glasses keeping the card $25$ cm
away. Ten years later, he reads his farewell letter with
the same glasses but he has 'to keep the letter $50$ cm
away. What power of lens should he now use ?

On the $50^{th}$ birthday, he reads the card at a distance 25cm using a glass of +2.5D. $\\$ Ten years later, his near point must have changed. $\\$ So after ten years, $\\$ u = – 50 cm, $\qquad$ $f = \frac{1}{2.5D}=$ 0.4m = 40 cm $\qquad$ v = near point $\\$ Now, $\frac{1}{v}-\frac{1}{u} = \frac{1}{f} \quad \Rightarrow \frac{1}{v} = \frac{1}{u} + \frac{1}{f} = \frac{1}{-50} + \frac{1}{40} = \frac{1}{200}$ $\\$ So, near point = v = 200cm $\\$ To read the farewell letter at a distance of 25 cm, $\\$ U = – 25 cm $\\$ For lens formula, $\\$ $\frac{1}{v}- \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{f} = \frac{1}{200} - \frac{1}{-25} = \frac{1}{200}+\frac{1}{25} = \frac{9}{200} \\ \Rightarrow f = \frac{200}{9} cm = \frac{2}{9} cm$ $\\$ $\Rightarrow$ Power of the lens $= \frac{1}{f} = \frac{9}{2} = 4.5 D$ $\\$ $\therefore$ He has to use a lens of power +4.5D.

**19.** A normal eye has retina $2$ cm behind the eye-lens. What
is the power of the eye-lens when the eye is (a) fully
relaxed, (b) most strained ?

Since, the retina is 2 cm behind the eye-lens $\\$ v = 2cm $\\$ (a) When the eye-lens is fully relaxed $\\$ u = $\infty$ , v = 2cm = 0.02 m $\\$ $\Rightarrow \frac{1}{f} = \frac{1}{v}-\frac{1}{u} = \frac{1}{0.02}-\frac{1}{\infty} = 50 D$ $\\$ So, in this condition power of the eye-lens is 50D $\\$ (b) When the eye-lens is most strained, $\\$ u = – 25 cm = – 0.25 m, $\qquad \quad$ v = +2 cm = +0.02 m $\\$ $\Rightarrow \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{0.02}- \frac{1}{-0.25} = 50+4 = 54 D$ $\\$ In this condition power of the eye lens is 54D.$\\$

**20.** The near point and the far point of a child are at $10$ cm
and $100$ cm. If the retina is $2.0$ cm behind the eye-lens,
what is the range of the power of the eye-lens ?

The child has near point and far point 10 cm and 100 cm respectively. $\\$ Since, the retina is 2 cm behind the eye-lens, v = 2cm $\\$ For near point u = – 10 cm = – 0.1 m, $\qquad$ v = 2 cm = 0.02 m $\\$ So, $\frac{1}{f_{near}} = \frac{1}{v}-\frac{1}{u} = \frac{1}{0.02}- \frac{1}{-0.1} = $ 50 + 10 = 60D $\\$ For far point, u = – 100 cm = – 1 m, $\qquad$ v = 2 cm = 0.02 m $\\$ So, $\frac{1}{f_{far}} = \frac{1}{v}-\frac{1}{u} = \frac{1}{0.02}- \frac{1}{-1} = $ 50 + 1 = 51D $\\$ So, the rage of power of the eye-lens is +60D to +51D $\\$

**21.** A nearsighted person cannot see beyond $25$ cm.
Assuming that the separation of the glass from the eye
is $1$ cm, find the power of lens needed to see distant
objects.

For the near sighted person, $\\$ v = distance of image from glass $\\$ = distance of image from eye – separation between glass and eye $\\$ = 25 cm – 1cm = 24 cm = 0.24m $\\$ So, for the glass, u = $\infty$ and v = – 24 cm = –0.24m $\\$ So, $\frac{1}{f} = \frac{1}{v} -\frac{1}{u} = \frac{1}{-0.24} - \frac{1}{\infty} = -4.2 D$

**22.** A person has near point at $100$ cm. What power of lens
is needed to read at $20$ cm if he/she uses (a) contact lens,
(b) spectacles having glasses $2.0$ cm separated from the
eyes?

The person has near point 100 cm. It is needed to read at a distance of 20cm. $\\$ (a) When contact lens is used, $\\$ u = – 20 cm = – 0.2m, $\qquad$ v = – 100 cm = –1 m $\\$ So, $\frac{1}{f}= \frac{1}{v}-\frac{1}{u} = \frac{1}{-1}-\frac{1}{-0.2} = -1+5=+4D$ $\\$ (b) When spectacles are used,$\\$ u = – (20 – 2) = – 18 cm = – 0.18m, $\qquad$ v = – 100 cm = –1 m $\\$ So, $\frac{1}{f}= \frac{1}{v}-\frac{1}{u} = \frac{1}{-1}-\frac{1}{-0.18} =$ – 1 + 5.55 = + 4.5D $\\$

**23.** A lady uses + $1.5$ D glasses to have normal vision from
25 cm onwards. She uses a $20$ D lens as a simple
microscope to see an object. Find the maximum
magnifying power if she uses the microscope (a) together
with her glass (b) without the glass. Do the answers
suggest that an object can be more clearly seen through
a microscope without using the correcting glasses ?

The lady uses +1.5D glasses to have normal vision at 25 cm. $\\$ So, with the glasses, her least distance of clear vision = D = 25 cm $\\$ Focal length of the glasses $= \frac{1}{1.5}m = \frac{100}{1.5}cm$ $\\$ So, without the glasses her least distance of distinct vision should be more $\\$ If, u = – 25cm, $\qquad$ $f=\frac{100}{1.5} cm$ $\\$ Now,$\frac{1}{v}-\frac{1}{u} = \frac{1}{f} = \frac{1.5}{100}-\frac{1}{25} = \frac{1.5-4}{100} = \frac{-2.5}{100} \qquad $ $\\$ $\Rightarrow$ v = – 40cm = near point without glasses. $\\$ Focal length of magnifying glass $= \frac{1}{20}m =$ 0.05m = 5 cm = f $\\$ (a) The maximum magnifying power with glasses $\\$ $m = 1+\frac{D}{f} = 1+\frac{25}{5}=6 \qquad $ [ $\because$ D = 25cm] $\\$ (b) Without the glasses, D = 40cm $\\$ So, m $=1+ \frac{D}{f} = 1+ \frac{40}{5} = 9$ $\\$

**24.** A lady cannot see objects closer than $40$ cm from the
left eye and closer than $100$ cm from the right eye. While
on a mountaineering trip, she is lost from her team. She
tries to make an astronomical telescope from her reading
glasses to look for her teammates. (a) Which glass should
she use as the eyepiece ? (b) What magnification can she
get with relaxed eye ?

The lady can not see objects closer than 40 cm from the left eye and 100 cm from the right eye. $\\$ For the left glass lens, $\\$ v = – 40 cm, $\qquad$ u = – 25 cm $\\$ $\therefore \frac{1}{f}= \frac{1}{v}-\frac{1}{u} = \frac{1}{-40}- \frac{1}{-25}= \frac{1}{25}- \frac{1}{40} = \frac{3}{200}$ $\\$ $\Rightarrow f=\frac{200}{3} cm$ $\\$ For the right glass lens, $\\$ v = – 100 cm, $\qquad$ u = – 25 cm $\\$ $ \frac{1}{f}= \frac{1}{v}-\frac{1}{u} = \frac{1}{-100}- \frac{1}{-25}= \frac{1}{25}- \frac{1}{100} = \frac{3}{100}$ $\\$ $\Rightarrow f=\frac{100}{3} cm$ $\\$ (a) For an astronomical telescope, the eye piece lens should have smaller focal length. So, she should use the right lens $ ( f = \frac{100}{3}cm)$ as the eye piece lens. $\\$ (b) With relaxed eye, (normal adjustment) $\\$ $f_0=\frac{200}{3}cm, \qquad f_e = \frac{100}{3} cm$ $\\$ magnification = m $= \frac{f_0}{f_e} = \frac{(200/3)}{(100/3)} = 2$