Physics 5 Marks Questions

$\text{f}_0=\frac{1}{25\text{ diopter}}=0.04\text{m}=4\text{cm,}$ $\text{f}_\text{e}=\frac{1}{5\text{ diopter}}=0.2\text{m}=20\text{cm}$

$\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{f}_\text{e}}$

$\Rightarrow\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{f}_\text{e}}$

$\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

v₀ = 30 - (11.11) = 18.89cm

Now, for the objective lens,

v₀ = +18.89cm (because real image is produced)

f₀ = 4cm

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

$\text{m}=-\frac{\text{v}_\text{o}}{\text{u}_\text{o}}\Big[1+\frac{\text{D}}{\text{f}_\text{e}}\Big]=\frac{18.89}{-5.07}\Big[1+\frac{25}{20}\Big]$

$=3.7225\times2.25=8.376$

$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\Rightarrow\frac{1}{\text{v}}-\frac{1}{-30}=\frac{1}{15}$

$\Rightarrow\frac{1}{\text{v}}=\frac{1}{30}$

$\Rightarrow \text{v}=+30\text{cm}$

(on the opposite side of the object)

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

f₀ = = 1cm, f_e = 5cm, u₀ = 0.5cm, v_e = 30cm

For the objective lens, u₀ = – 0.5cm, f₀ = 1cm.

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 1cm 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 = u₀ – v₀ = 6 – 1 = 5cm

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 $=\frac{1}{20}\text{m}=0.05\text{m}=5\text{cm}=\text{f}$

1. The maximum magnifying power with glasses

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

2. Without the glasses, D = 40cm.

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

$\text{f}_0=\frac{1}{25\text{D}}=0.04\text{m}=4\text{cm}$

$\text{f}_\text{e}=\frac{1}{20\text{D}}=0.05\text{m}=5\text{cm}$

tube length = 25cm (normal adjustment)

1. The instrument must be a microscope as f₀ < f_e
2. 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 = v₀ = 25 – 5 = 20cm

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 = 25cm]

$=-\frac{20}{-5}\times\frac{25}{5}=20$

f₀ = 1cm, f_e = 6cm, D = 24cm

For the eye piece, v_e = -24cm, f_e = 6cm

Now, $\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{f}_\text{e}}$

$\Rightarrow\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{f}_\text{e}}\Rightarrow-\Big[\frac{1}{24}+\frac{1}{6}\Big]=-\frac{5}{24}$

$\Rightarrow \text{u}_\text{e}=-4.8\text{cm}$

1. When the separation between objective and eye piece is 9.8cm, the image distance for the objective lens must be (9.8) - (4.8) = 5.0cm

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,

$\text{m}=\frac{\text{v}_0}{\text{u}_\text{0}}\Big[1+\frac{\text{D}}{\text{f}}\Big]=\frac{-5}{-1.25}\Big[1+\frac{24}{6}\Big]=4\times5=20$

2.  When the separation is 11.8cm,

v₀ = 11.8 - 4.8 = 7.0cm, f₀ = 1cm

$\Rightarrow\frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}=\frac{1}{7}-\frac{1}{1}=-\frac{6}{7}$

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.

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.

$\text{f}_0=\frac{1}{20\text{D}}=0.05\text{m}=5\text{cm},$ $\text{f}_\text{e}=\frac{1}{10\text{D}}=0.1\text{m}=10\text{cm.}$

D = 25cm, separation between objective & eyepiece= 20cm

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, v₀ = -25cm, f_e = 10cm

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 $=\frac{0.22}{5.5}\text{mm}=0.04\text{mm}$

So, v_o + f_e = 6.5cm ...(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 = 25cm]

$\Rightarrow100\text{f}_\text{e}=-(1-2\text{v}_0)\times25$

$\Rightarrow2\text{v}_0-4\text{f}_\text{e}=1 $ ...(2)

Solving equation (1) and (2) we can get,

V₀ = 4.5cm and f_e = 2cm

So, the focal length of the eye piece is 2cm.

v = 2cm

1. 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

2. 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.

Magnifying power = m = 50, length of the tube = L = 102cm

Let f₀ 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 f0 from equation (1) in (2), we get,

f₀ + f_e = 102 ⇒ 51f_e = 102 ⇒ f_e = 2cm = 0.02m

So, f₀ = 100cm = 1m

$\therefore$ Power of the objective lens$=\frac{1}{\text{f}_0}=1\text{D}$

And Power of the eye piece lens$=\frac{1}{\text{f}_\text{e}}=\frac{1}{0.02}=50\text{D}$