Ray Optics and Optical Test

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Ray Optics and Optical Test - 71

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Weekly Quiz Competition

Question 1
1 / -0

A convex lens of focal length $$f$$ produces a real image 3 times the size of an object, then distance of the object from the lens is

A
$$-\dfrac{2f} { 3}$$

B
$$-\dfrac{\mathrm { f }} { 2}$$

C
$$-\dfrac{\mathrm { f } }{ 4}$$

D
$$-\dfrac{\mathrm {4 f }} { 3}$$

Solution
Question 2
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A ray of light is incident normally on one face of $$30^{ \circ }-60^{ \circ }-90^{ \circ }$$ prism of refractive index 5/3 immersed in water of refractive index 4/3 as shown in figure .

A
The exit angle $${ \theta }_{ 2 }$$ of the ray is $${ sin }^{ -1 }$$ (5/8)

B
The exit angle $${ \theta }_{ 2 }$$ of the ray is $${ sin }^{ -1 }$$ $$(5/4\sqrt { 3 } )$$

C
Total internal reflection at point P ceases if the refractive index of water is increased to $$5/4\sqrt { 3 } $$ by dissolving some substance .

D
Total internal reflection at point P ceases if the refractive index of water is increase to 5/6 by dissolving some substance .

Solution

Given,
The refractive index pf prism is $${\mu _p} = \dfrac{5}{3}$$ and the refractive index of water is $${\mu _w} = \dfrac{4}{3}$$.

From the figure it can be seen that $${\theta _1} = 60^\circ $$

Applying Snell’s law

$${\mu _p}\sin 30^\circ = {\mu _w}\sin {\theta _2}$$

$$\dfrac{5}{3}\sin 30^\circ = \dfrac{4}{3}\sin {\theta _2}$$

$$\dfrac{5}{3} \times \dfrac{1}{2} = \dfrac{4}{3}\sin {\theta _2}$$

$$\sin {\theta _2} = \dfrac{3}{4} \times \dfrac{5}{3} \times \dfrac{1}{2}$$

$$\sin {\theta _2} = \dfrac{5}{8}$$

$${\theta _2} = {\sin ^{ - 1}}\left( {\dfrac{5}{8}} \right)$$
Hence the exit angle of the ray is $${\theta _2} = {\sin ^{ - 1}}\left( {\dfrac{5}{8}} \right)$$ .
Question 3
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Consider slabs of three media A,B and C arranged as shown in figure R.I of A is 1.5 and that of C is 1.4. If the number of waves in A is equal to the number of waves in the combination of B and C then refractive index of B is:

A
1.4

B
1.5

C
1.6

D
1.7

Solution

If $${\lambda _0}$$ be the wavelength outside the slab, then we can say that

$${\mu _{A,B,C}} = \dfrac{{{\lambda _0}}}{{{\lambda _{A,B,C}}}}$$

$$\therefore {\lambda _A} = \dfrac{{{\lambda _0}}}{{{\mu _A}}}$$ $${\lambda _B} = \dfrac{{{\lambda _0}}}{{{\mu _B}}}$$ $${\lambda _C} = \dfrac{{{\lambda _0}}}{{{\mu _C}}}$$

Number of waves$$ = \dfrac{d}{\lambda }$$

From the figure we have the distances

$${d_A} = 3x$$ $${d_B} = x$$ $${d_C} = 2x$$

$$\therefore \dfrac{{{d_A}}}{{{\lambda _A}}} = \dfrac{{{d_B}}}{{{\lambda _B}}} + \dfrac{{{d_C}}}{{{\lambda _C}}}$$

$$ \Rightarrow \dfrac{{3x}}{{{\lambda _0}/{\mu _A}}} = \dfrac{x}{{{\lambda _0}/{\mu _B}}} + \dfrac{{2x}}{{{\lambda _0}/{\mu _C}}}$$

$$ \Rightarrow 3{\mu _A} = {\mu _B} + 2{\mu _C}$$

$$ \Rightarrow {\mu _B} = 3{\mu _A} - 2{\mu _C}$$

$$ \Rightarrow {\mu _B} = 3 \times 1.5 - 2 \times 1.4 = 1.7$$

Hence the correct answer is option (D).
Question 4
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Two thin prisms are combined to form an achromatic combination.For prism I, $$ A=4^{0}, \mu_{R}=1.35, \mu_{Y}=1.40, \mu_{\nu}=1.42 . $$ For prism II $$ \mu_{R}=1.7, \mu_{\gamma}=1.8 $$ and $$ \mu_{R}=1.9 . $$ Find the prism angle of prism II and the net mean deviation.

A
$$ 0.44^{\circ} $$

B
$$ 0.45^{\circ} $$

C
$$ 0.48^{\circ} $$

D
$$ 0.46^{\circ} $$

Solution

The correct option is (C).

Given,
The reflecting angle of prism I is $$A = 4^\circ $$
Now for aromatic combination
$$\left( {{\mu _{v1}} - {\mu _{R1}}} \right){A_1} = \left( {{\mu _{v2}} - {\mu _{R2}}} \right){A_2}$$
$$\left( {1.42 - 1.35} \right) \times 4^\circ = \left( {1.9 - 1.7} \right){A_2}$$
$${A_2} = \dfrac{{\left( {1.42 - 1.35} \right) \times 4^\circ }}{{\left( {1.9 - 1.7} \right)}}$$
$${A_2} = 1.4^\circ $$
Now the net minimum deviation is given as
$${\delta _1} - {\delta _2} = \left( {{\mu _{Y1}} - 1} \right){A_1} - \left( {{\mu _{Y2}} - 1} \right){A_2}$$
$${\delta _1} - {\delta _2} = \left( {1.40 - 1} \right) \times 4^\circ - \left( {1.8 - 1} \right) \times 1.4^\circ $$
$${\delta _1} - {\delta _2} = 0.48^\circ $$.
Question 5
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A ray of light is incident on a glass at an angle of $$60^o$$. What would be the refractive index of glass, if reflected and refracted rays are perpendicular to each other ?

A
$$\sqrt{3}$$

B
$$\dfrac{3}{2}$$

C
$$\dfrac{\sqrt{3}}{2}$$

D
$$\dfrac{1}{2}$$

Solution

The incident ray $$1$$ will be reflected at an equal angle as reflected ray $$2.$$
By laws of reflection, $$i=r=60^o$$
The refracted ray $$3$$ is perpendicular to reflected ray $$2.$$
So by geometry, the angle of refraction $$r'=30^o$$
Now by Snell's law,
$$\mu_{air} \ sin \ i= \mu_{glass} \ sin \ r'$$
$$(1) \times sin \ 60^o= \mu_{glass} \times sin \ 30^o$$
$$\dfrac{\sqrt3}{2}=\mu_{glass} \times \dfrac{1}{2}$$
$$\mu_{glass}=\sqrt3$$
Option A is the answer.
Question 6
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A glass slab of thickness 12 mm is placed on a table. The lower surface of the slab has a black spot. At what depth from the upper surface will the spot appear when viewed from above? (Refractive index of glass =1.5)

A
$$2 mm$$

B
$$4 mm$$

C
$$6 mm$$

D
$$8 mm$$

Solution

Here it is given,

Read depth $$D = 12mm$$

Refractive index $$\mu = 1.5$$

Let $$d$$ be the apparent depth then the formula for apparent depth is given by

$$\mu = \dfrac{D}{d}$$

$$ \Rightarrow d = \dfrac{D}{\mu } = \dfrac{{12}}{{1.5}} = 8mm$$

Hence the correct answer is option (A).
Question 7
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A right-angled prism is set to deviate a ray of light through $$90^o$$ . If the same prism is to be used for deviating through $$180^o$$ , the prism must be turned from the previous set position through the angle :

A
$$45^o$$

B
$$90^o$$

C
$$135^o$$

D
$$180^o$$

Solution

From the figure we can see that the right angled prism needs to be rotated by $${45^\circ }$$.

Hence the correct answer is option (A).
Question 8
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An object is seen through a glass slab of thickness 36 cm and refractive index 3/2. The observer, object and the slab are dipped in water (n= 4/3). the shift produced in the position of the object is

A
12 cm

B
4 cm

C
cannot be calculated

D
9/2 cm

Solution

Shift produced in the position of the object-
$$s = \left( {1 - \frac{{{\mu _{medium}}}}{{{\mu _{slab}}}}} \right) \times t$$
Here,
$${\mu _{medium}}$$ is the refractive index of the surrounding medium
$${\mu _{slab}}$$ is the refractive index of the slab material
$$t$$ is the thickness of the slab
We put all the values and get-
$$s = \left( {1 - \frac{{(4/3)}}{{(3/2)}}} \right) \times 36{\text{ cm}}$$
$$s = \left( {1 - \frac{8}{9}} \right) \times 36{\text{ cm}}$$
$$s = \left( {\frac{1}{9}} \right) \times 36{\text{ cm}}$$
$$s = 4{\text{ cm}}$$

So, option (B) Is correct.
Question 9
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A parallel air film is formed between two glass plates. If the film thickness is $$0.45\times 10^{-6} m$$, the film will be best reflector for visible light in the neighbourhood of

A
$$9000\overset {\circ}{A}$$

B
$$6750\overset {\circ}{A}$$

C
$$6000\overset {\circ}{A}$$

D
$$4500\overset {\circ}{A}$$
Question 10
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A student measures the focal length of a convex lens by putting an object pin at a distance 'u' from the lens and measured the distance 'v' of the image pin. The graph between 'u' and 'v' plotted by the student should look like

A

B

C

D