3 Marks Question

Question 1013 Marks

The focal lengths of a convex lens for red, yellow and violet rays are 100cm, 98cm and 96cm respectively. Find the dispersive power of the material of the lens.

Answer

The focal length of a lens is given by
$\frac{1}{\text{f}}=(\mu-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$
$\Rightarrow(\mu-1)=\frac{1}{\text{f}}\times\frac{1}{\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)}=\frac{\text{k}}{\text{f}}\ ...(1)$
So, $\mu_\text{r}-1=\frac{\text{K}}{100}\ ...(2)$
$\mu_\text{y}-1=\frac{\text{K}}{98}\ ...(3)$
And $\mu_\text{v}-1=\frac{\text{K}}{96}\ ...(4)$
So, Dispersive power $=\omega=\frac{\mu_\text{v}-\mu_\text{r}}{\mu_\text{y}-1}=\frac{(\mu_\text{v}-1)-(\mu_\text{r}-1)}{(\mu_\text{y}-1)}=\frac{\frac{\text{K}}{96}-\frac{\text{K}}{100}}{\frac{\text{K}}{98}}$ $=\frac{98\times4}{9600}=0.0408$

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Question 1023 Marks

The magnifying power of a simple microscope is given by $1+\frac{\text{D}}{\text{f}},$ where D is the least distance for clear vision. For farsighted persons, D is greater than the usual. Does it mean that the magnifying power of a simple microscope is greater for a farsighted person as compared to a normal person? Does it mean that a farsighted person can see an insect more clearly under a microscope than a normal person?

Answer

The magnifying power of a simple microscope depends on the ratio $\frac{\text{D}}{\text{f}}$ for a farsighted person. Here, D for a farsighted person is greater than that for a normal person, but the value of f remains the same. Therefore, the magnifying power of a simple microscope is greater for a farsighted person compared to that for a person with normal vision. Also, a farsighted person can see the insect more clearly under the microscope than a person with normal vision.

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Question 1033 Marks

A small piece of wood is floating on the surface of a 2.5m deep lake. Where does the shadow form on the bottom when the sun is just setting? Refractive index of water $=\frac{4}{3}.$

Answer

Height of the lake = 2.5m

When the sun is just setting, $\theta$ is approximately = 90°

$\therefore \ \frac{\sin\text{i}}{\sin\text{r}}=\frac{\mu_2}{\mu_1}\Rightarrow\frac{1}{\sin\text{r}}=\frac{\frac{4}{3}}{1}\Rightarrow\sin\text{r}=\frac{3}{4}\Rightarrow\text{r}=49^{\circ}$

As shown in the figure, $\frac{\text{x}}{2.5}=\tan\text{r}=1.15$

$\Rightarrow\text{x}=2.5\times1.15=2.8\text{m}.$

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Question 1043 Marks

A light ray falling at an angle of 45^° with the surface of a clean slab of ice of thickness 1m is refracted into it at an angle of 30^°. Calculate the time taken by the light rays, to cross the slab. Speed of light in vacuum = 3 × 10⁸m/s

Answer

We know, $\frac{\sin\text{i}}{\sin\text{r}}=\frac{3\times10^8}{\text{v}}=\frac{\sin45^{\circ}}{\sin30^{\circ}}=\sqrt{2}$

$\Rightarrow\text{v}=\frac{3\times10^8}{\sqrt{2}}\text{m}/\text{sec}$

Distance travelled by light in the slab is,

$\text{x}=\frac{1\text{m}}{\cos30^{\circ}}=\frac{2}{\sqrt{3}}\text{m}$

So, time taken $=\frac{2\times\sqrt{2}}{\sqrt{3}\times3\times10^8}=0.54\times10^{-8}=5.4\times10^{-9}\text{sec}.$

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Question 1053 Marks

A lens forms a real image of an object. The distance from the object to the lens is u cm and the distance of the image from the lens is v cm. The given graphs shows variation of v with u.

What is the nature of the lens?
Using the graph, find the focal length of this lens.

Answer

The lens is convex because image formed by the lens is real.

Focal length: From graph when u = -40cm, v = 15cm.

$\therefore \frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$ (for a lens)

$=\frac{1}{15}+\frac{1}{40}=\frac{40+15}{40\times15}$

$\text{f}=\frac{40\times15}{55}=\frac{120}{11}\approx11\text{cm.}$

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Question 1063 Marks

A concave mirror having a radius of curvature 40cm is placed in front of an illuminated point source at a distance of 30cm from it. Find the location of the image.

Answer

u = -30cm, R = -40cm

From the mirror equation,

$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{2}{\text{R}}$

$\Rightarrow \frac{1}{\text{v}}=\frac{2}{\text{R}}-\frac{1}{\text{u}}=\frac{2}{-30}-\frac{1}{-30}=-\frac{1}{60}$

or, v = -60cm

So, the image will be formed at a distance of 60cm in front of the mirror.

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Question 1073 Marks

A simple microscope has a magnifying power of 3.0 when the image is formed at the near point (25cm) of a normal eye.

What is its focal length?
What will be its magnifying power if the image is formed at infinity?

Answer

The simple microscope has, m = 3, when image is formed at D = 25cm

$\text{m}=1+\frac{\text{D}}{\text{f}}\Rightarrow3=1+\frac{25}{\text{f}}$

$\Rightarrow\text{f}=\frac{25}{2}=12.5\text{cm}$

When the image is formed at infinity (normal adjustment)

Magnifying power $=\frac{\text{D}}{\text{f}}=\frac{25}{12.5}= 2.0$

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Question 1083 Marks

Two lenses of power 10D and -5D are placed in contact.

Calculate the power of lens combination.
Where should an object be held from the lens, so as to obtain a virtual image of magnification 2?

Answer

Given P₁ = 10 D,

P₂ = -5D

Power of Combination, P = p₁ + P₂ = 10D - 5D = 5D

Focal length (Convergent lens) $\text{f}=\frac{1}{\text{P}}=\frac{1}{5}\text{m}$

$=0.20\text{m}=20\text{cm}$

(Convergent lens)

Magnification $\text{m}=\frac{\text{v}}{\text{u}}=+2$

$\Rightarrow\text{v}=2\text{u}$

From lens formula (u is negative) $\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$

$\frac{1}{20}=\frac{1}{2\text{u}}-\frac{1}{\text{u}}$

$\Rightarrow-\frac{1}{2\text{u}}=\frac{1}{20}$

$\Rightarrow\text{u}=-10\text{cm}.$

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Physics 3 Marks Question