Ray Optics and Optical Test

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Ray Optics and Optical Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

Colour of the sky is blue due to

A
Scattering of light

B
Total internal reflection

C
Total emission

D
None of the above

Solution

The molecules of air and other fine particles in the atmosphere have size smaller than the wavelength of visible light. These are more effective in scattering light of shorter wavelengths at the blue end than light of longer wavelengths at the red end. Thus, when sunlight passes through the atmosphere, the fine particles in air scatter the blue colour (shorter wavelengths) more strongly than red. The scattered blue light enters our eyes.
Question 2
1 / -0

A simple magnifying lens is used in such a way that an image is formed at $$25\ cm$$ away from the eye. In order to have $$10$$ time magnification, the focal length of the lens should be

A
$$5\ cm$$

B
$$2\ cm$$

C
$$25\ mm$$

D
$$0.1\ mm$$

Solution

As we know,
$$m=1+\dfrac{D}{f_e}\Rightarrow 10=1+\dfrac{25}{f_e}\Rightarrow f_e=\dfrac{25}{9}\simeq 25\ mm$$
Question 3
1 / -0

Which of the following event is not related to refraction of light ?

A
The bottom of water filled bowl appears raised.

B
Appearance of sun before sun rising and after sunset

C
Formation of image by mirror

D
Twinkling of stars

Solution
Question 4
1 / -0

The apreture of astronomical telescope is large because .

A
to remove spherical defect

B
for high limit of revolution

C
to anlarge observation area

D
for less dispersion

Solution

The aperture of astronomical is large to remove spherical aberration
Question 5
1 / -0

Choose the correct alternative:
A ray of light make an angle of $$50^{\circ} $$ with the surface $$S_{1}$$ of the glass slab. Its angle of incidence will be__________

A
$$50^{\circ} $$

B
$$40^{\circ} $$

C
$$140^{\circ} $$

D
$$0^{\circ} $$

Solution
Question 6
1 / -0

A uniform, horizontal beam of light is incident upon a prism as shown in the fig. The prism is in the shape of a quarter cylinder of radius R $$=$$ 5cm, and has a refractive index $$\dfrac{2}{\sqrt{3}}$$. A patch on the table for a distance x from the cylinder is unilluminated. The value of x is

A
$$2.5$$ cm

B
$$5$$ cm

C
$$5 \sqrt{3}$$cm

D
$$10$$ cm

Solution

Figure-(I) shows how for rays nearer to base are refracted but as rays move away from base, they fall nearer and after a certain distance above base, all rays are internally reflected.
So, we have to find the distance x where there is no light. As on increasing height rays fall nearer to the prism, so we have to find where will the ray for critical angle of incidence will intersect base line. And inside that point there will be no illumination as for larger angle of incidence , rays are internally reflected.

From figure-2, it is clear that $$i=C$$=critical angle and refracted ray for this goes perpendicular to normal which is radius $$OA$$ here. Refracted ray is $$AB$$.

$$i=C=\sin^{_1}\dfrac{1}{\mu}=\sin^{-1}\dfrac{\sqrt{3}}{2}=60^{o}$$

From $$\Delta OAB:-$$

$$\cos i=\dfrac{OA}{OB}=\dfrac{R}{R+x}$$

$$\implies\cos 60^{o}= \dfrac{1}{2}=\dfrac{5}{5+x}$$

$$\implies x=5cm$$

Answer-(B)
Question 7
1 / -0

The refracting angle of a prism $$60^{\mathrm{o}}$$.The refractive index of the material of the prism is $$\sqrt{\dfrac{7}{3}}$$. The limiting angle of incidence of a ray that will be transmitted through the prism in this case will be :

A
$$30^{0}$$

B
$$45^{\mathrm{o}}$$

C
$$40^{\mathrm{o}}$$

D
$$50^{\mathrm{o}}$$

Solution

$$\begin{array}{l}\text { Given } A=60^{\circ}, \mu=\sqrt{\frac{7}{3}} \\\text { we know that } \gamma_{1}+r_{2}=A \\\Rightarrow r_{1}+r_{2}=60^{\circ}\end{array}$$
$$\begin{array}{l}\text { Snell's law at first refraction } \\\qquad \begin{array}{l}\sin i=\mu \sin \gamma_{1} \\\operatorname{Sin}\gamma_{1}=\sqrt{\frac{3}{7}} \sin i\end{array} \\\text { Snell's law at second refraction }\end{array}$$
$$\begin{aligned}\mu \sin r_{2} &=1 \\ & \Rightarrow \sin r_{2}=\sqrt{\frac{3}{7}} \\ r_{2}=\sin ^{-1}\left(\sqrt{\frac{3}{7}}\right) \\ r_{2}=& 40.89^{\circ} \Rightarrow r_{1}=19.1066^{\circ}\end{aligned}$$

$$\begin{aligned}\Rightarrow\operatorname{Sin} r_{1} &=0.327 \\\sin i &=\sqrt{\frac{7}{3}} \sin r_{1} \\\sin i &=1 / 2 \\ i &=30^{\circ}\end{aligned} $$
Question 8
1 / -0

A ray of light enters at grazing angle of incidence into an assembly of three isosceles right-angled prisms having refractive indices $$\mu_{1}=\sqrt{2}, \mu_{2}=\sqrt{x}$$ and $$\mu_{3}=\sqrt{3}$$ . If finally emergent light ray also emerges at grazing angle then calculate x :

A
5

B
3

C
2

D
1

Solution

Apply Snell's law on various surfaces one by one :

1 sin $$90^{\circ}$$ $$=\mu_{1}$$ sin $$r_{1} \Rightarrow$$ sin $$r_{1} = \dfrac{1}{\sqrt{2}} \Rightarrow r_{1}=45^{\circ}$$

We have the incident angle for the second interface given using the relation sin$$i_2$$=cos$$r_1$$
Thus we get

$$\mu_{1}$$ cos $$r_{1}$$= $$\mu_{2}$$ sin $$r_{2} \Rightarrow $$ sin $$r_{2}$$ = $$\dfrac{1}{\mu_{2}}$$(cos$$r_2$$=$$\dfrac{1}{\sqrt2}$$)

$$\mu_{2}$$ cos $$r_{2}=\mu_{3}$$ sin $$r_{3}$$ $$\Rightarrow$$ sin $$r_{3}$$ $$=\dfrac{\mu_{2}\sqrt{1-sin^{2}r_{2}}}{\sqrt{3}}$$

$$\mu_{3}$$ cos $$r_{3}=1$$ = $$\dfrac{\mu_{2}^{2}-1}{\sqrt{3}}$$

sin $$^{2} r_{3}$$ + cos $$^{2} r_{3}=1$$

$$\Rightarrow \dfrac{\mu_{2}^{2}-1}{3}+\dfrac{1}{3}=1 \Rightarrow {\mu_{2}}^{2}=3\Rightarrow \mu_{2}=\sqrt{3}$$

Thus we get $$x=3.$$
Question 9
1 / -0

Refractive indices of 2 different media with separating boundary at the diagonal of rectangular glass slab are shown. Total angle of deviation of the ray as shown in the figure, when it emerges in air is :

A
$$120^{\circ}$$

B
$$90^{\circ}$$

C
$$60^{\circ}$$

D
$$45^{\circ}$$

E
$$30^{\circ}$$

Solution

Angle of incident at the boundary = 60
By snells law
$$\Rightarrow \sqrt{3}\sin 60^{\circ}=1.5 \sin r$$
$$\Rightarrow r=90^{\circ}$$
Thus the angle of deviation as seen in the figure is given as $$90^o-60^o=30^o$$
Question 10
1 / -0

A thin convex lens made of glass of $$\mu =$$ 1.5 has refracting surfaces of radii of curvature 10 cm each. The left space of lens contains air and the right space is filled with water of refractive index $$\dfrac{4}{3}$$. A parallel beam of light is incident on it. The position of the image is :

A
$$16 cm$$

B
$$20 cm$$

C
$$24cm$$

D
$$10 cm$$

Solution

Data: $${ \mu }_{ a }=1,{ \mu }_{ g }=1.5,{ \mu }_{ w }=1.33,{ R }_{ 1 }=10cm,{ R }_{ 2 }=-10cm,u=-\infty $$

For left surface,
$$\dfrac{\mu_g}{v_1}-\dfrac{\mu_a}{u} = \dfrac{\mu_g-\mu_a}{R_1}$$

For right surface,
$$\dfrac{\mu_w}{v}-\dfrac{\mu_g}{v_1} = \dfrac{\mu_w-\mu_g}{R_2}$$

From above two,
$$ \dfrac { { \mu }_{ w } }{ v } -\dfrac { { \mu }_{ a } }{ u } =\dfrac { { \mu }_{ g }-{ \mu }_{ a } }{ { R }_{ 1 } } +\dfrac { { \mu }_{ w }-{ \mu }_{ g } }{ { R }_{ 2 } } $$

or, $$ \dfrac { 1.33 }{ v } =\dfrac { 1.5-1 }{ 10 } -\dfrac { 1.33-1 }{ 10 } $$

$$v=\dfrac { 1.33\times 10 }{ 0.66 } =20\ cm$$