3 Marks Question

Question 13 Marks

The diameter of the sun is 1.4 × 10⁹m and its distance from the earth is 1.5 × 10¹¹m. Find the radius of the image of the sun formed by a lens of focal length 20cm.

Answer

u = -1.5 × 10¹¹ m; v = +20 × 10^-2 m

Since, f is very small compared to u, distance is taken as $\infty.$

So, image will be formed at focus.

⇒ v = +20 × 10^-2m

$\therefore$ We know, $\text{m}=\frac{\text{v}}{\text{u}}=\frac{\text{h}_{\text{image}}}{\text{h}_{\text{object}}}$

$\Rightarrow\frac{20\times10^{-2}}{1.5\times10^{11}}=\frac{\text{D}_{\text{image}}}{1.4\times10^9}$

$\Rightarrow\text{D}_{\text{image}}=1.86\text{mm}$

So, radius $=\frac{\text{D}_{\text{image}}}{2}=0.93\text{mm}.$

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Question 23 Marks

A light ray, going through a prism with the angle of prism 60°, is found to deviate by 30°. What limit on the refractive index can be put from these data?

Answer

Given $\text{A}=60^{\circ}$ and $\mu=30^{\circ}$

We know that,

$\mu=\frac{\sin\Big(\frac{\text{A}+\delta_{\text{m}}}{2}\Big)}{\frac{\sin\text{A}}{2}}=\frac{\sin\frac{60^{\circ}+\delta_{\text{m}}}{2}}{\sin30^{\circ}}=2\sin\frac{60^{\circ}+\delta_{\text{m}}}{2}$

Since, one ray has been found out which has deviated by 30°, the angle of minimum deviation should be either equal or less than 30°. (It can not be more than 30°).

So, $\mu\leq2\sin\frac{60^{\circ}+\delta_{\text{m}}}{2}$ $\big($because $\mu$ will be more if $\delta_{\mu}$ will be more$\big)$

or, $\mu\leq2\times\frac{1}{\sqrt{2}}$

or, $\mu\leq\sqrt{2}.$

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Question 33 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 43 Marks

If a piece of paper is placed at the position of a virtual image of a strong light source, will the paper burn after sufficient time? What happens if the image is real? What happens if the image is real but the source is virtual?

Answer

No, Yes, Yes.
Real Image: Rays are really getting focussed at point P so intensity is high.

Virtual Image: Ray appears to come from point P and no real concentration of ray at P.

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Question 53 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 63 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 73 Marks

A man uses a concave mirror for shaving. He keeps his face at a distance of 25cm from the mirror and gets an image which is 1.4 times enlarged. Find the focal length of the mirror.

Answer

$\text{u}=-25\text{cm}$

$\text{m}=\frac{\text{A}'\text{B}'}{\text{AB}}=-\frac{\text{v}}{\text{u}}\Rightarrow1.4=-\Big(\frac{\text{v}}{-25}\Big)\Rightarrow\frac{14}{10}=\frac{\text{v}}{25}$

$\Rightarrow\text{v}=\frac{25\times14}{10}=35\text{cm}$

Now, $\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\Rightarrow\frac{1}{\text{f}}=\frac{1}{35}-\Big(\frac{1}{-25}\Big)$

$\Rightarrow\frac{5-7}{175}=\frac{-2}{175}$

$\Rightarrow\text{f}=-87.5\text{cm}$

So, focal length of the concave mirror is 87.5cm.

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Question 83 Marks

A light ray is incident at an angle of 45° with the normal to a $\sqrt{2}\text{cm}$ thick plate $(\mu = 2.0)$. Find the shift in the path of the light as it emerges out from the plate.

Answer

Applying snell's law,

$\frac{\sin \text{i}}{\sin \text{r}}=\frac{1}{\mu}$

$\Rightarrow \frac{\sin \text{i}}{\sin \text{r}}=\frac{2}{1}$

As shown in the figure,

$\frac{\sin45^{\circ}}{\sin\text{r}}=\frac{2}{1}\Rightarrow\sin\text{r}=\frac{\sin45^{\circ}}{2}=\frac{1}{2\sqrt{2}}\Rightarrow\text{r}=21^{\circ}$

Here, BD = shift in path = AB sin 24°

$=0.406\times\text{AB}=\frac{\text{AE}}{\cos21^{\circ}}\times0.406=0.62\text{cm}.$

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Question 93 Marks

Find the angle of deviation suffered by the light ray shown in figure. The refractive index $\mu=1.5$ for the prism material.

Answer

Given, $\mu=1.5$

And angle of prism $=4^{\circ}$

$\therefore \ \mu=\frac{\sin\big(\frac{\text{A}+\delta_{\text{m}}}{2}\big)}{\frac{\sin\text{A}}{2}}=\frac{\big(\frac{\text{A}+\delta_{\text{m}}}{2}\big)}{\big(\frac{\text{A}}{2}\big)}$ $\big($for small angle $\sin\theta=\theta\big)$

$\Rightarrow\mu=\frac{\text{A}+\delta_{\text{m}}}{2}\Rightarrow1.5=\frac{4^{\circ}+\delta_{\text{m}}}{4^{\circ}}$

$\Rightarrow\delta_{\text{m}}=4^{\circ}\times(1.5)-4^{\circ}=2^{\circ}$

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Question 103 Marks

A point object is placed on the principal axis of a convex lens (f = 15cm) at a distance of 30cm from it. A glass plate $(\mu=1.50)$ of thickness 1cm is placed on the other side of the lens perpendicular to the axis. Locate the image of the point object.

Answer

For the lens, f = 15cm, u = 30cm

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

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

The image is formed at 30cm of right side due to lens only.

Again, shift due to glass slab is,

$=\Delta\text{t}=\Big(1-\frac{1}{15}\Big)1 \ \big[\text{Since,} \ \mu_{\text{g}=1.5 \ \text{and} \ \text{t}=1\text{cm}}\big]$

$=1-\Big(\frac{2}{3}\Big)=0.33\text{cm}$

$\therefore$ The image will be formed at 30 + 0.33 = 30.33cm from the lens on right side.

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Question 113 Marks

Find the angle of minimum deviation for an equilateral prism made of a material of refractive index 1.732. What is the angle of incidence for this deviation?

Answer

Refractive index $(\mu)$ of the material from which prism is made = 1.732

We know refractive index is given by:

$\mu=\frac{\Big[\frac{\delta_{\text{min}}+\text{A}}{2}\Big]}{\sin\Big[\frac{\text{A}}{2}\Big]}$

Where $\delta_{\text{min}}$ is the angleof minimum deviation and A is the angle of prism = 60º

$\Rightarrow 1.732\times \sin(30^\circ)=\sin\Big(\frac{\delta_{\text{min}}+60^\circ}{2}\Big)$

$\Rightarrow \frac{1.732}{2}=\sin\Big(\frac{\delta_{\text{min}}+60^\circ}{2}\Big)$

$\Rightarrow \Big(\frac{\delta_{\text{min}}+60^\circ}{20}\Big)=60^\circ$

$\delta_{\text{min}}=60^\circ$

$\delta_{\text{min}}=2\text{i}-\text{A}$

$2\text{i}=120^\circ$

$\text{i}=60^\circ$

Hence, the required angle of deviation is 60°.

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Question 123 Marks

Locate the image of the point P as seen by the eye in the figure.

Answer

The presence of air medium in between the sheets does not affect the shift. The shift will be due to 3 sheets of different refractive index other than air.

$\Delta \text{t}=\Big[1-\frac{1}{\mu_1}\Big]\text{t}_1+\Big[1-\frac{1}{\mu_2}\Big]\text{t}_2+\Big[1-\frac{1}{\mu_3}\Big]\text{t}_3$

$=\Big(1-\frac{1}{1.2}\Big)(0.2)+\Big(1-\frac{1}{13}\Big)(0.3)+\Big(1-\frac{1}{14}\Big)(0.4)$

$=0.2\text{cm}$ above point P.

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Question 133 Marks

A 5.0 diopter lens forms a virtual image which is 4 times the object placed perpendicularly on the principal axis of the lens. Find the distance of the object from the lens.

Answer

Given, P = 5 diopter (convex lens)

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

Since, a virtual image is formed, u and v both are negative.Given, $\frac{\text{v}}{\text{u}}=4$

$\Rightarrow\text{v}=4\text{u} \ ...(1)$

From lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\Rightarrow\frac{1}{\text{f}}=\frac{1}{4\text{u}}-\frac{1}{\text{u}}\Rightarrow\frac{1}{20}=\frac{1-4}{4\text{u}}=-\frac{3}{4\text{u}}$

$\Rightarrow\text{u}=-15\text{cm}$

$\therefore$ Object is placed 15cm away from the lens.

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Question 143 Marks

One end of a cylindrical glass rod $(\mu=1.5)$ of radius 1.0cm is rounded in the shape of a hemisphere. The rod is immersed in water $\Big(\mu=\frac{4}{3}\Big)$ and an object is placed in the water along the axis of the rod at a distance of 8.0cm from the rounded edge. Locate the image of the object.

Answer

Radius of the cylindrical glass tube = 1cm

We know, $\frac{\mu_2}{\text{v}}-\frac{\mu_1}{\text{u}}=\frac{\mu_2-\mu_1}{\text{R}}$

Here, $\text{u}=-8\text{cm}, \ \mu_2=\frac{3}{2},\mu_1=\frac{4}{3},\text{R}=+1\text{cm}$

So, $\frac{3}{2\text{v}}+\frac{4}{3\times8}\Rightarrow\frac{3}{2\text{v}}+\frac{1}{6}=\frac{1}{6}$

$ \ \text{v}=\infty$

$\therefore$ The image will be formed at infinity.

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Question 153 Marks

A concave mirror has a focal length of 20cm. Find the position or positions of an object for which the imagesize is double of the object-size.

Answer

For the concave mirror,

$\text{f}=-20\text{cm}, \ \text{M}=-\frac{\text{v}}{\text{u}}=2$

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

1^st case:	2^nd case:
$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$	$\frac{-1}{2\text{u}}-\frac{1}{\text{u}}=-\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{2\text{u}}-\frac{1}{\text{u}}=-\frac{1}{\text{f}}$	$\Rightarrow\frac{3}{2\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\text{u}=\frac{\text{f}}{2}=10\text{cm}$	$\Rightarrow\text{u}=\frac{3\text{f}}{2}=30\text{cm}$

$\therefore$ The positions are 10cm or 30cm from the concave mirror.

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Question 163 Marks

Solve the previous problem if the paperweight is inverted at its place so that the spherical surface touches the paper.

Answer

Thickness of glass $=3\text{cm}, \ \mu_{\text{g}}=1.5$

Image shif $=3\Big(1-\frac{1}{1.5}\Big)$

[Treating it as a simple refraction problem because the upper surface is flat and the spherical surface is in contact with the object]

$=3\times\frac{0.5}{1.5}=1\text{cm}.$

The image will appear 1 cm above the point P.

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Question 173 Marks

A 3cm tall object is placed at a distance of 7.5cm from a convex mirror of focal length 6cm. Find the location, size and nature of the image.

Answer

Given AB = 3cm, u = -7.5cm, f = 6cm.

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

Putting values according to sign conventions,

$\frac{1}{\text{v}}=\frac{1}{6}-\frac{1}{-7.5}=\frac{3}{10}$

$\Rightarrow\text{v}=\frac{10}{3}\text{cm}$

$\therefore$ magnification $=\text{m}=-\frac{\text{v}}{\text{u}}=\frac{10}{7.5\times3}$

$\Rightarrow\frac{\text{A}'\text{B}'}{\text{A}\text{B}}=\frac{10}{7.5\times3}\Rightarrow\text{A}'\text{B}'=\frac{100}{72}=\frac{4}{3}=1.33\text{cm}$

$\therefore$ Image will form at a distance of $\frac{10}{3}\text{cm}$ From the pole and image is 1.33cm (virtual and erect).

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Question 183 Marks

A vessel contains water upto a height of 20cm and above it an oil upto another 20cm. The refractive indices of the water and the oil are 1.33 and 1.30 respectively. Find the apparent depth of the vessel when viewed from above.

Answer

Shift due to water $\Delta\text{t}_\text{w}=\Big(1-\frac{1}{\mu}\Big)\text{d}\Big(1-\frac{1}{1.33}\Big)20=5\text{cm}$
Shift due to oil, $\Delta\text{t}_0=\Big(1-\frac{1}{1.33}\Big)20=4.6\text{cm}$
Total shift $\Delta\text{t}=5+4.6=9.6\text{cm}$
Apparent depth = 40 - (9.6) = 30.4cm below the surface.

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Question 193 Marks

An object P is focused by a microscope M. A glass slab of thickness 2.1cm is introduced between P and M. If the refractive index of the slab is 1.5, by what distance should the microscope be shifted to focus the object again?

Answer

The thickness of the glass is $\text{d} = 2.1\text{cm}$ and $\mu=1.5$ Shift due to the glass slab

$\Delta\text{T}=\Big(1-\frac{1}{\mu}\Big)\text{d}$

$=\frac{1}{3}(2.1)=0.7\text{cm}$

So, the microscope should be shifted 0.70cm to focus the object again.

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Question 203 Marks

A narrow beam of light passes through a slab obliquely and is then received by an eye. The index of refraction of the material in the slab fluctuates slowly with time. How will it appear to the eye? The twinkling of stars has a similar explanation.

Answer

The image position shifts because of variation in refractive index. Thus it appears to be twinkling to the eyes.

When $\mu$ changes $\alpha$ changes and also x changes so ray appear to come from different position.

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Question 213 Marks

Locate the image formed by refraction in the situation shown in figure.

Answer

$\mu_1=1, \ \mu_2=1.5,$

R = 20cm (Radius of curvature), u = -25cm

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

$\Rightarrow\frac{1.5}{\text{v}}=\frac{0.5}{20}-\frac{1}{25}=\frac{1}{40}-\frac{1}{25}=\frac{-3}{200}$

$\Rightarrow \ \text{v}=200\times0.5=-100\text{cm}.$

So, the image is 100cm from (P) the surface on the side of S.

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Question 223 Marks

A concave mirror of radius R is kept on a horizontal. Water $\big($refractive index $=\mu\big)$ is poured into it upto a height h. Where should an object be placed so that its image is formed on itself?

Answer

Let the object be placed at a height x above the surface of the water.

The apparent position of the object with respect to mirror should be at the centre of curvature so that the image is formed at the same position.

Since, $\frac{\text{Real depth}}{\text{Apparent depth}}=\frac{1}{\mu}$ (with respect to mirror)

Now, $\frac{\text{x}}{\text{R}-\text{h}}=\frac{1}{\mu}$

$\Rightarrow\text{x}=\frac{\text{R}-\text{h}}{\mu}$

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Question 233 Marks

A particle goes in a circle of radius 2.0cm. A concave mirror of focal length 20cm is placed with its principal axis passing through the centre of the circle and perpendicular to its plane. The distance between the pole of the mirror and the centre of the circle is 30cm. Calculate the radius of the circle formed by the image.

Answer

$\text{u}=-30\text{cm}, \ \text{f}=-20\text{cm}$

We know, $\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$

$\Rightarrow\frac{1}{\text{v}}+\Big(-\frac{1}{30}\Big)=\Big(-\frac{1}{20}\Big)\Rightarrow\text{v}=-60\text{cm}$

Image of the circle is formed at a distance 60cm in front of the mirror.

$\therefore \ \text{m}=-\frac{\text{v}}{\text{u}}=\frac{\text{R}_{\text{image}}}{\text{R}_{\text{object}}}\Rightarrow\frac{-60}{-30}=\frac{\text{R}_{\text{image}}}{2}$

$\Rightarrow\text{R}_{\text{image}}=4\text{cm}$

Radius of image of the circle is 4cm.

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Question 243 Marks

A converging lens and a diverging mirror are placed at a separation of 15cm. The focal length of the lens is 25cm and that of the mirror is 40cm. Where should a point source be placed between the lens and the mirror so that the light, after getting reflected by the mirror and then getting transmitted by the lens, comes out parallel to the principal axis?

Answer

If the image in the mirror will form at the focus of the converging lens, then after transmission through the lens the rays of light will go parallel.

Let the object is at a distance x cm from the mirror

$\therefore$ u = -x cm; v = 25 - 15 = 10cm (because focal length of lens = 25cm)

f = 40cm

$\Rightarrow\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{x}}=\frac{1}{10}-\frac{1}{40}$

$\Rightarrow\text{x}=\frac{400}{30}=\frac{40}{3}$

$\therefore$ The object is at distance $\Big(15-\frac{40}{3}\Big)=\frac{5}{3}=1.67\text{cm}$ from the lens.

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Question 253 Marks

A convex lens produces a double size real image when an object is placed at a distance of 18cm from it. Where should the object be placed to produce a triple size real image?

Answer

A real image is formed. So, magnification m = -2(inverted image)

$\therefore\frac{\text{v}}{\text{u}}=-2\Rightarrow\text{v}=-2\text{u}=(-2)(-18)=36$

From lens formula, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{36}-\frac{1}{-18}=\frac{1}{\text{f}}$

$\Rightarrow\text{f}=12\text{cm}$

Now, for triple sized image $\text{m}=-3=\Big(\frac{\text{v}}{\text{u}}\Big)$

$\therefore \ \frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{-3\text{u}}-\frac{1}{\text{u}}=\frac{1}{12}$

$\Rightarrow3\text{u}=-48\Rightarrow\text{u}=-16\text{cm}$

So, object should be placed 16cm from lens.

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Question 263 Marks

A candle flame 1.6cm high is imaged in a ball bearing of diameter 0.4cm. If the ball bearing is 20cm away from the flame, find the location and the height of the image.

Answer

Height of the object AB = 1.6cm

Diameter of the ball bearing = d = 0.4cm

⇒ R = 0.2cm

Given, u = 20cm

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

Putting the values according to sign conventions $\frac{1}{-20}+\frac{1}{\text{v}}=\frac{2}{0.2}$

$\Rightarrow\frac{1}{\text{v}}=\frac{1}{20}+10=\frac{201}{20}\Rightarrow\text{v}=0.1\text{cm}=1\text{mm}$ inside the ball bearing.

Magnification $=\text{m}=\frac{\text{A}'\text{B}'}{\text{AB}}=-\frac{\text{v}}{\text{u}}=-\frac{0.1}{-20}=\frac{1}{200}$

$\Rightarrow\text{A}'\text{B}'=\frac{\text{A}\text{B}}{200}=\frac{16}{200}=+0.008\text{cm}=+0.8\text{mm}.$

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Question 273 Marks

A pole of length 1.00m stands half dipped in a swimming pool with water level 50.0 cm higher than the bed. The refractive index of water is 1.33 and sunlight is coming at an angle of 45° with the vertical. Find the length of the shadow of the pole on the bed.

Answer

Shadow length $= \text{BA}' = \text{BD} + \text{A}'\text{D} = 0.5 + 0.5 \tan \text{r}$

Now, $1.33=\frac{\sin45^{\circ}}{\sin\text{r}}\Rightarrow\sin\text{r}=0.53$

$\Rightarrow\cos\text{r}=\sqrt{1-\sin^2\text{r}}=\sqrt{1-(0.53)^2}=0.85$

So, $\tan \text{r} = 0.6235$

So, shadow length = (0.5)(1 + 0.6235) = 81.2cm

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