Case study (4 Marks)

Question 14 Marks

In motor vehicles, a convex mirror is attached near the driver's seat to give him the view of the traffic behind. What is the special function of this convex mirror which a plane mirror can not do?

Answer

Field of view is large.
Field of view is more for mirror 2

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Question 24 Marks

Suppose you are inside the water in a swimming pool near an edge. A friend is standing on the edge. Do you find your friend taller or shorter than his usual height?

Answer

Taller, When view from outside water, the height is $x_2-x_1$; but when viewed from inside water, the height is $\mu(\text{x}_2-\text{x}_1).$

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Question 34 Marks

A narrow pencil of parallel light is incident normally on a solid transparent sphere of radius r. What should be the refractive index if the pencil is to he focused.

At the surface of the sphere.
At the centre of the sphere.

Answer

For the refraction at convex surface A.$\mu=-\infty, \ \mu_1=1, \ \mu_2=?$

When focused on the surface, v = 2r, R = r

So, $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$
$\Rightarrow\frac{\mu_2}{2\text{r}}=\frac{\mu_2-1}{\text{r}}$
$\Rightarrow \frac{\mu_2}{2\text{r}}-\Big(\frac{1}{-\infty}\Big)=\frac{\mu_2-1}{\text{r}}$
$\Rightarrow\mu_2=2\mu_2-2\Rightarrow\mu_2=2$

When focused at centre, $u = r_1$, R = r

So, $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$
$\Rightarrow\frac{\mu_2}{\text{R}}=\frac{\mu_2-1}{\text{r}}\Rightarrow\mu_2=\mu_2-1.$
This is not possible.
So, it cannot focus at the centre

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Question 44 Marks

A slide projector has to project a 35mm slide (35mm × 23mm) on a 2m × 2m screen at a distance of 10m from the lens. What should be the focal length of the lens in the projector?

Answer

For the projector the magnification required is given by$\text{m}=\frac{\text{v}}{\text{u}}=\frac{200}{3.5}\Rightarrow\text{u}=17.5\text{cm}$
[35 mm > 23mm, so the magnification is calculated taking object size 35mm] Now, from lens formula,$\Rightarrow\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{v}}-\frac{1}{-\text{u}}=\frac{1}{\text{f}}\Rightarrow \frac{1}{1000}+\frac{1}{17.5}=\frac{1}{\text{f}}$
$\Rightarrow\text{f}=17.19\text{cm}.$

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Question 54 Marks

Consider the situation described in the previous problem. Where should a point source be placed on the principal axis so that the two images form at the same place?

Answer

Let the source be placed at a distance ‘x’ from the lens as shown, so that images formed by both coincide. For the lens, $\frac{1}{\text{v}_\ell}-\frac{1}{-\text{x}}=\frac{1}{15}\Rightarrow\text{v}_\ell=\frac{15\text{x}}{\text{x}-15} \ ...(1)$ For the mirror, $\text{u}=-(50-\text{x}), \ \text{f}=-10\text{cm}$ So, $\frac{1}{\text{v}_\text{m}}+\frac{1}{-(50-\text{x})}=-\frac{1}{10}$$\Rightarrow\frac{1}{\text{v}_\text{m}}=\frac{1}{-(50-\text{x})}-\frac{1}{10}$
$\text{v}_\text{m}=\frac{10(50-\text{x})}{\text{x}-40} \ ...(2)$
Since the lens and mirror are 50cm apart,$\text{v}_\ell-\text{v}_{\text{m}}=50\Rightarrow\frac{15\text{x}}{\text{x}-15}-\frac{10(50-\text{x})}{(\text{x}-40)}=50$
$\Rightarrow\text{x}=30\text{cm}.$
So, the source should be placed 30cm from the lens.

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Question 64 Marks

A small object is embedded in a glass sphere $(\mu=1.5)$ of radius 5.0cm at a distance 1.5cm left to the centre. Locate the image of the object as seen by an observer standing.

To the left of the sphere.
To the right of the sphere.

Answer

Image seen from left:

u = (5 - 15) = -3.5cm

R = -5cm

$\therefore \ \frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$

$\Rightarrow\frac{1}{\text{v}}+\frac{1.5}{3.5}=-\frac{1-1.5}{5}$

$\Rightarrow\frac{1}{\text{v}}=\frac{1}{10}-\frac{3}{7}\Rightarrow\text{v}=\frac{-70}{23}=-3\text{cm}$ (inside the sphere).

$\Rightarrow$ Image will be formed, 2cm left to centre.

Image seen from right:

u = -(5 + 1.5) = -6.5cm

R = -5cm

$\therefore \ \frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$

$\Rightarrow\frac{1}{\text{v}}+\frac{1.5}{6.5}=\frac{1-1.5}{-5}$

$\Rightarrow\frac{1}{\text{v}}=\frac{1}{10}-\frac{3}{13}$

$\Rightarrow \frac{13-30}{130}$

$\Rightarrow\text{v}=\frac{-130}{17}=-7.65\text{cm}$ (inside the sphere).

$\Rightarrow$ Image will be formed, 2.65cm left to centre.

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Question 74 Marks

A container contains water upto a height of 20cm and there is a point source at the centre of the bottom of the container. A rubber ring of radius r floats centrally on the water. The ceiling of the room is 2.0m above the water surface.

Find the radius of the shadow of the ring formed on the ceiling if r = 15cm.
Find the maximum value of r for which the shadow of the ring is formed on the ceiling. Refractive index of water $\frac{4}{3}.$

Answer

As shown in the figure, $\sin\text{i}=\frac{15}{25}$

So, $\frac{\sin\text{i}}{\sin\text{r}}=\frac{1}{\mu}=\frac{3}{4}$

$\Rightarrow\sin\text{r}=\frac{4}{5}$

Again, $\frac{\text{x}}{2}=\tan\text{r}$ (from figure)

So, $\sin\text{r}=\frac{\tan\text{r}}{\sqrt{1+\tan^2\text{r}}}=\frac{\frac{\text{x}}{2}}{\sqrt{1+\frac{\text{x}^2}{4}}}$

$\Rightarrow\frac{\text{x}}{\sqrt{{4}+\text{x}^2}}=\frac{4}{5}$

$\Rightarrow25\text{x}^2=16(4+\text{x}^2)\Rightarrow9\text{x}^2=64\Rightarrow\text{x}=\frac{8}{3}\text{m}$

$\therefore$ Total radius of shadow $=\frac{8}{3}+0.15=2.81\text{m}$

For maximum size of the ring, i = critical angle = C

Let, R = maximum radius

$\Rightarrow\sin\theta_\text{C}=\frac{\sin\theta_\text{C}}{\sin\text{r}}=\frac{\sin\text{R}}{\sqrt{20^2+\sin\text{R}^2}}=\frac{3}{4}$(since, sin r = 1)

$\Rightarrow16\text{R}^2=9\text{R}^2+9\times400$

$\Rightarrow7\text{R}^2=9\times400$

$\Rightarrow\text{R}=22.67\text{cm.}$

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Physics Case study (4 Marks)

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