5 Marks Questions

Question 15 Marks

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 $20D$ objective and $10D$ 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?

Answer

For the given compound microscope $\text{f}_0=\frac{1}{20\text{D}}=0.05\text{m}=5\ \text{cm},$
$\text{f}_\text{e}=\frac{1}{10\text{D}}=0.1\text{m}=10\ \text{cm.}$
$D = 25\ cm,$ 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, $\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{f}_\text{e}}=\frac{1}{-25}-\frac{1}{10}$
$=-\Big[\frac{2+5}{50}\Big]$
$\Rightarrow\text{u}_\text{e}=-\frac{50}{7}\text{cm}$
So, the image distance for the objective lens should be,
$\text{v}_0=20-\frac{50}{7}=\frac{90}{7}\text{cm}$
Now, for the objective lens,
$\frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}=\frac{7}{90}-\frac{1}{5}=-\frac{11}{90}$
$\Rightarrow\text{u}_\text{o}=-\frac{90}{11}\text{cm}$
So, the maximum magnifying power is given by,
$\text{m}=\frac{-\text{V}_0}{\text{u}_0}\Big[1+\frac{\text{D}}{\text{f}_\text{e}}\Big]$
$=\frac{\big(\frac{90}{7}\big)}{\big(-\frac{90}{11}\big)}\Big[1+\frac{25}{10}\Big]$
$=\frac{11}{7}\times3.5=5.5$
Thus, minimum separation eye can distinguish $=\frac{0.22}{5.5}\ \text{mm}=0.04\ \text{mm}$

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Question 25 Marks

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.

Answer

For the give compound microscope$, f_0 = 0.5\ cm,$ 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$, v_o + f_e = 6.5\ cm ...(1)$
Again, magnifying power $\text{m}=\frac{\text{v}_0}{\text{u}_0}\times\frac{\text{D}}{\text{f}_\text{e}}\ [$for normal adjustment$]$
$\Rightarrow\text{m}=\Big[1-\frac{\text{V}_0}{\text{f}_0}\Big]\frac{\text{D}}{\text{f}_\text{e}}$
$\Rightarrow100=-\Big[1-\frac{\text{V}_0}{0.5}\Big]\times\frac{25}{\text{f}_\text{e}}\ [$Taking $D = 25\ cm]$
$\Rightarrow100\text{f}_\text{e}=-(1-2\text{v}_0)\times25$
$\Rightarrow2\text{v}_0-4\text{f}_\text{e}=1 ...(2)$
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 $2\ cm.$

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Question 35 Marks

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 $25\ cm$.

Answer

For the given compound microscope $\text{f}_0=\frac{1}{25\text{ diopter}}=0.04\text{m}=4\text{cm,} \text{f}_\text{e}=\frac{1}{5\text{ diopter}}=0.2\text{m}=20\text{cm}$
$D = 25\ cm,$ separation between objective and eyepiece $= 30\ cm$
The magnifying power is maximuwm hen the image is formed by the eye piece at least distance of clear vision i.e. $D = 25\ cm$ for the eye piece$, v_e= -25\ cm, f_e = 20\ cm$
For lens formula, $\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{f}_\text{e}}$$\Rightarrow\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{f}_\text{e}}$
$\Rightarrow\frac{1}{\text{u}_\text{e}}\frac{1}{-25}-\frac{1}{20} $
$\Rightarrow\text{u}_\text{e}=11.11\ \text{cm}$
So, for the objective lens, the image distance should be $v_0 = 30 - (11.11) = 18.89\ cm$
Now, for the objective lens$, v_0 = +18.89\ cm\ ($because real image is produced$) f_0 = 4\ cm$
So, $\frac{1}{\text{u}_\text{o}}=\frac{1}{\text{v}_\text{o}}-\frac{1}{\text{f}_\text{o}}$
$\Rightarrow \frac{1}{18.89}-\frac{1}{4}$
$=0.053-0.25=-0.197$
$\Rightarrow\text{u}_\text{o}=-5.07\text{cm}$
So, the maximum magnificent power is given by $\text{m}=-\frac{\text{v}_\text{o}}{\text{u}_\text{o}}\Big[1+\frac{\text{D}}{\text{f}_\text{e}}\Big]=\frac{18.89}{-5.07}\Big[1+\frac{25}{20}\Big]$
$=3.7225\times2.25=8.376$

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Question 45 Marks

An object is placed at a distance of 30cm from a converging lens of focal length 15cm. A normal eye (near point 25cm, far point infinity) is placed close to the lens on the other side.

Can the eye see the object clearly?
What should be the minimum separation between the lens and the eye so that the eye can clearly see the object?
Can a diverging lens, placed in contact with the converging lens, help in seeing the object clearly when the eye is close to the lens?

Answer

Object distance, u = -30cm Focal length, f = 15cm Image distance, v = ? The lens formula is given by$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{v}}-\frac{1}{-30}=\frac{1}{15}$
$\Rightarrow\frac{1}{\text{v}}=\frac{1}{30}$
$\Rightarrow \text{v}=+30\text{cm}$
(on the opposite side of the object)

No, the eye placed close to the lens cannot see the object clearly.
The eye should be 30cm away from the lens to see the object clearly.
The diverging lens will always form an image at a large distance from the eye this image cannot be seen through the human eye.

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Question 55 Marks

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?

Answer

Given that$,f_0 = = 1\ cm, f_e = 5\ cm, u_0 = 0.5\ cm, v_e = 30\ cm$
For the objective lens$, u_0 = – 0.5\ cm, f_0 = 1\ cm.$
From lens formula.
$\frac{1}{\text{v}_0}-\frac{1}{\text{u}_0}=\frac{1}{\text{f}_0}$
$\Rightarrow\frac{1}{\text{v}_0}-\frac{1}{\text{u}_0}=\frac{1}{\text{f}_0}$
$\Rightarrow\frac{1}{\text{v}_0}=\frac{1}{\text{u}_0}+\frac{1}{\text{f}_0}$
$\Rightarrow\frac{1}{\text{v}_0}=\frac{1}{-0.5}+\frac{1}{1}=-1$
$\Rightarrow\text{v}_0=-1\text{cm}$
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,
$\frac{1}{\text{v}_0}-\frac{1}{\text{u}_0}=\frac{1}{\text{f}_0}$
$\Rightarrow \frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}$
$\Rightarrow\frac{1}{\text{u}_0}=\frac{1}{30}-\frac{1}{5}=\frac{-5}{30}=\frac{-1}{6}$
$\Rightarrow\text{u}_\text{o}=-6\text{cm}$
So, as shown in figure,
Separation between the lenses $= u_0 – v_0 = 6 – 1 = 5\ cm$

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Question 65 Marks

A lady uses +1.5D glasses to have normal vision from 25cm onwards. She uses a 20D lens as a simple microscope to see an object. Find the maximum magnifying power if she uses then microscope

Together with her glass
Without the glass. Do the answers suggest that an object can be more clearly seen through a microscope without using the correcting glasses?

Answer

The lady uses +1.5D glasses to have normal vision at 25cm.So, with the glasses, her least distance of clear vision = D = 25cm
Focal length of the glasses $=\frac{1}{1.5}\text{m}=\frac{100}{1.5}\text{cm}$
So, without the glasses her least distance of distinct vision should be more
If, $\text{u}=-25\text{cm},$
$\text{f}=\frac{100}{1.5}\text{cm}$
Now, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}=\frac{1.5}{100}-\frac{1}{25}=\frac{1.5{-4}}{100}=\frac{-2.5}{100}$
$\Rightarrow \text{v}=-40\text{cm}$ = near point without glasses
Focal length of magnifying glass $=\frac{1}{20}\text{m}=0.05\text{m}=5\text{cm}=\text{f}$

The maximum magnifying power with glasses

$\text{m} =1+\frac{\text{D}}{\text{f}}=1+\frac{25}{5}=6$

Without the glasses, D = 40cm.

So, $\text{m}=1+\frac{\text{D}}{\text{f}}=1+\frac{40}{5}=9$

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Question 75 Marks

An optical instrument used for angular magnification has a $25D$ objective and a $20D$ eyepiece. The tube length is $25\ cm$ when the eye is least strained.

Whether it is a microscope or a telescope?
What is the angular magnification produced?

Answer

The optical instrument has $\text{f}_0=\frac{1}{25\text{D}}=0.04\text{m}=4\text{cm}$
$\text{f}_\text{e}=\frac{1}{20\text{D}}=0.05\text{m}=5\text{cm}$
tube length $= 25\ cm\ ($normal adjustment$)$

The instrument must be a microscope as $f_0< f_e$
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.
$\frac{1}{\text{v}_0}-\frac{1}{\text{u}_0}=\frac{1}{\text{f}_0}$
$\Rightarrow\frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}=\frac{1}{20}-\frac{1}{4}$
$=\frac{-4}{20}=\frac{-1}{5}\Rightarrow\text{u}_0=-5\text{cm}$
So, angular magnification $\text{m}=-\frac{\text{v}_0}{\text{u}_0}\times\frac{\text{D}}{\text{f}_\text{e}}\ [$Taking $D = 25\ cm]$
$=-\frac{20}{-5}\times\frac{25}{5}=20$

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Question 85 Marks

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.

Answer

For the given compound microscope $f_0 = 1\ cm, f_e = 6\ cm, D = 24\ cm$
For the eye piece$, v_e= -24\ cm, f_e = 6\ cm$
Now, $\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{f}_\text{e}}$
$\Rightarrow\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{f}_\text{e}}$
$\Rightarrow-\Big[\frac{1}{24}+\frac{1}{6}\Big]=-\frac{5}{24}$
$\Rightarrow \text{u}_\text{e}=-4.8\ \text{cm}$

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, $\frac{1}{\text{v}_0}-\frac{1}{\text{u}_0}=\frac{1}{\text{f}_0}$
$\Rightarrow\frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}=\frac{1}{5}-\frac{1}{1}=-\frac{4}{5}$
$\Rightarrow\text{u}_0=-\frac{5}{4}=-1.25\text{cm}$
So, the magnifying power is given by,
$\text{m}=\frac{\text{v}_0}{\text{u}_\text{0}}\Big[1+\frac{\text{D}}{\text{f}}\Big]=\frac{-5}{-1.25}\Big[1+\frac{24}{6}\Big]=4\times5=20$

When the separation is $11.8\ cm,$

$v_0 = 11.8 - 4.8 = 7.0\ cm, f_0 = 1\ cm$
$\Rightarrow\frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}=\frac{1}{7}-\frac{1}{1}=-\frac{6}{7}$
So, $\text{m}=-\frac{\text{v}_0}{\text{u}_\text{o}}\Big[1+\frac{\text{D}}{\text{f}}\Big]$
$=\frac{-7}{-\Big(\frac{7}{6}\Big)}\Big[1+\frac{24}{6}\Big]=6\times5=30$
So, the range of magnifying power will be $20$ to $30.$

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Question 95 Marks

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

Answer

On the 50th 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,
$\text{u}=-50\text{cm},$
$\text{f}=\frac{1}{2.5\text{D}}=0.4\text{m}=40\text{cm}$
$\text{v}=\text{near point}$
Now, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow \frac{1}{\text{v}}=\frac{1}{\text{u}}+\frac{1}{\text{f}}=\frac{1}{-50}+\frac{1}{40}=\frac{1}{200}$
So, near point = v = 200cm
To read the farewell letter at a distance of 25cm,
U = –25cm
For lens formula,
$\Rightarrow\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{f}}=\frac{1}{200} -\frac{1}{-25}=\frac{1}{200}+\frac{1}{25}=\frac{9}{200}$
$\Rightarrow\text{f}=\frac{200}{9}\text{cm}=\frac{2}{9}\text{m}$
⇒ Power of the lens $=\frac{1}{\text{f}}=\frac{9}{2}=4.5\text{D.}$
$\therefore$ He has to use a lens of power +4.5D.

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Question 105 Marks

A normal eye has retina 2cm behind the eye-lens. What is the power of the eye-lens when the eye is

Fully relaxed,
Most strained?

Answer

Since, the retina is 2cm behind the eye-lensv = 2cm

When the eye-lens is fully relaxed

$\text{u}=\infty,$
$\text{v}=2\text{cm}=0.02\text{m}$
$\Rightarrow\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{0.02}-\frac{1}{\infty}=50\text{D}$
So, in this condition power of the eye-lens is 50D

When the eye-lens is most strained,

$\text{u}=-25\text{cm}=-0.25\text{m}$
$\text{v}=+2\text{cm}=+0.02\text{m}$
$\Rightarrow\frac{1}{\text{u}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{0.02}-\frac{1}{-0.25}=50+4=54\text{D}$
In this condition power of the eye lens is 54D.

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Question 115 Marks

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.

Answer

For the astronomical telescope in normal adjustment.
Magnifying power $= m = 50,$ length of the tube $= L = 102\ cm$
Let $f_0$ and fe be the focal length of objective and eye piece respectively.
$\text{m}=\frac{\text{f}_0}{\text{f}_\text{e}}=50$
$\Rightarrow\text{f}_0=50\text{f}_\text{e}\dots(1)$
and, $\text{L}=\text{f}_0+\text{f}_\text{e}=102\text{cm}\dots(2)$
Putting the value of $f_0$ from equation $(1)$ in $(2),$ we get,
$f_0 + f_e= 102$
$\Rightarrow 51f_e = 102 $
$\Rightarrow f_e= 2\ cm = 0.02m$
So$, f_0 = 100\ cm = 1m$
$\therefore$ Power of the objective lens$=\frac{1}{\text{f}_0}=1\text{D}$
And Power of the eye piece lens$=\frac{1}{\text{f}_\text{e}}=\frac{1}{0.02}=50\text{D}$

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