Ray Optics and Optical Test

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Ray Optics and Optical Test - 68

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Weekly Quiz Competition

Question 1
1 / -0

A object moving at a speed of $$5$$ m/s towards a concave mirror of focal length $$f=1$$m is at a distance of $$9$$m. The average speed of the image is?

A
$$\dfrac{1}{5}$$ m/s

B
$$\dfrac{1}{10}$$ m/s

C
$$\dfrac{4}{5}$$ m/s

D
$$\dfrac{2}{5}$$ m/s

Solution
Question 2
1 / -0

Two Nicol prisms are first crossed and then one of them is rotated through $$30^{0} $$ .The percentage of incident unpolarized light transmitted is

A
37.5

B
50

C
12.5

D
25.0

Solution
Question 3
1 / -0

A convex lens of focal length $$f$$ is placed somewhere between an object and a screen. The distance between the object and the screen is $$x$$. If the magnification is $$m$$, focal length of the lens is given as :

A
$$\dfrac { m x } { ( m + 1 ) ^ { 2 } }$$

B
$$\dfrac { ( m - 1 ) ^ { 2 } x } { m }$$

C
$$\dfrac { m x } { ( m - 1 ) ^ { 2 } }$$

D
$$\dfrac { ( m + 1 ) ^ { 2 } } { m } x$$

Solution

Given,

Focal length $$=f$$

$$v+u=x....(1)$$

magnification $$=m$$

magnification, $$m=\cfrac{v}{u}$$

$$v=um.....(2)$$

Put (2) in (1)

$$um+u=x$$

$$\Rightarrow$$ $$u=\cfrac{x}{m+1}.....(3)$$

from lens formula,

$$\cfrac { 1 }{ f } =\cfrac { 1 }{ v } -\cfrac { 1 }{ u } $$

but $$u=-u$$

$$\cfrac { 1 }{ f } =\cfrac { 1 }{ v } +\cfrac { 1 }{ u } $$

$$=\cfrac{1}{um}+\cfrac{1}{u}$$

$$\cfrac{1}{f}=\cfrac{1+m}{um}....(4)$$

put (3) in (4)

$$\Rightarrow$$ $$\cfrac{1}{f}=\cfrac{{(1+m)}^{2}}{xm}$$

$$\Rightarrow$$ $$f=\cfrac{xm}{{(1+m)}^{2}}$$
Question 4
1 / -0

Rays from an object immersed in water $$( \mu = 1.33 )$$ traverse a spherical air bubble of radius R. If the object is located far away from the bubble, its image as seen by the observer located on the other side of the bubble will be

A
virtual, erect and diminished

B
real, inverted and magnified

C
virtual, erect and magnified

D
real, inverted and diminished

Solution

$$\textbf{Given:-}$$ Radius of Spherical air bubble $$= R$$
$$ \mu = 1.33$$

$$\textbf{Solution:-} $$

Case - 1

$$\dfrac{1}{v} - \dfrac{\dfrac{4}{3}}{-\infty } = \dfrac{1 - \dfrac{4}{3}}{R} = - \dfrac{1}{3R} $$

$$\Rightarrow V = -3R$$

$$m_{1} = \dfrac{-\dfrac{3R}{1}}{-\dfrac{\dfrac{\infty }{4}}{3}} = +0$$

Case - 2

$$\dfrac{4}{3v} - \dfrac{1}{-5R} = \dfrac{\dfrac{4}{3} - 1}{-R}$$

$$\dfrac{4}{3v} + \dfrac{1}{5R} = -\dfrac{1}{3R}$$

$$\Rightarrow \dfrac{4}{3v} = -\dfrac{1}{3R} - \dfrac{1}{5R} = -\dfrac{8}{15R}$$

$$v = - \dfrac{5R}{2}$$

$$m_{2} = \dfrac{\dfrac{5R}{2(4/3)}}{- \dfrac{5R}{1}} = + \dfrac{3}{8}$$

$$m_{1}m_{2} = +0$$

image is virtual, erect and diminished

$$\textbf{Hence the correct option is A}$$
Question 5
1 / -0

Consider a ray of light incident from air onto a slab of glass ( reflective index n) of width d, at an angle $$\theta $$. The phase difference between the ray of reflected by the top surface of the glass and bottom surface is

A
$$\frac{4\pi d}{\lambda }\left ( 1-\frac{1}{n^{2}sin^{2}\theta } \right )^{1/2}+\pi $$

B
$$\frac{4\pi d}{\lambda }\left ( 1-\frac{1}{n^{2}sin^{2}\theta } \right )^{1/2} $$

C
$$\dfrac{4\pi d}{\lambda}\left (1+\dfrac{1}{n^{2}sin^{2}\theta }\right)^{1/2}$$

D
$$\dfrac{4\pi d}{\lambda}\left(1+\dfrac{1}{n^{2}sin^{2}\theta}\right)^{1/2}-\pi$$

Solution
Question 6
1 / -0

A mark is made on the bottom of a vessel and over this mark, a glass slab to thickness 3.5 cm and refractive index $$\frac{7}{4}$$ is placed. Now water (refractive index, $$\frac{4}{3}$$) is poured into the vessel so that the surface water is 8 cm above the upper surface of the slab. Looking down of normally through the water, the apparent depth of the mark below the surface of water will be :

A
6.33 cm

B
7.5 cm

C
8 cm

D
10 cm

Solution

Refractive index = real depth / apparent depth.

$$Apparent \ depth= \dfrac{8}{ \dfrac{ 4}{3} } + \dfrac{3.5}{ \dfrac{ 7}{4} } = 6 + 2 = 8 cm$$
Question 7
1 / -0

Monochromatic light is incident on a glass prism of angle $$A$$. If the refractive index of the material of the prism is $$\mu$$, a ray, incident at an angle $$\theta$$, on the face $$AB$$ would get transmitted through the face $$AC$$ of the prism provided.q

A
$$\theta > \sin^{-1} \left [\mu \sin \left (A - \sin^{-1} \left (\dfrac {1}{\mu}\right )\right )\right ]$$

B
$$\theta < \sin^{-1} \left [\mu \sin \left (A - \sin^{-1} \left (\dfrac {1}{\mu}\right )\right )\right ]$$

C
$$\theta > \cos^{-1} \left [\mu \sin \left (A + \sin \left (\dfrac {1}{\mu}\right )\right )\right ]$$

D
$$\theta < \cos^{-1} \left [\mu \sin \left (A + \sin \left (\dfrac {1}{\mu}\right )\right )\right ]$$

Solution
Question 8
1 / -0

Shown in the figure, a conical container of half -apex angle $${ 37 }^{ 0 }$$ filled with certain quantities of kerosene and water. The force exerted by the water on the kerosene is approximately. ( Take atmosphere pressure $${ 10 }^{ 0 }$$ pa)

A
$$3\times { 10 }^{ 7 }N$$

B
$$4\times { 10 }^{ 7 }N$$

C
$$2\times { 10 }^{ 7 }N$$

D
$$5\times { 10 }^{ 7 }N$$
Question 9
1 / -0

A ray of light is incident on the surface of separating two transparent medium at an angle $$45^0$$ and is refracted in medium at an angle $$30^0$$. Velocity of light in the medium will be...

A
$$3.8 \times 10^8 \ m/s$$

B
$$3.38 \times 10^8 \ m/s$$

C
$$2.12 \times 10^8 \ m/s$$

D
$$1.5 \times 10^8 \ m/s$$

Solution
Question 10
1 / -0

A convex lens (refractive index $$\mu=1.5$$) has a power P. If it is immersed in a liquid ($$\mu $$=4/3), then its power will become will become/remain

A
P

B
$$\dfrac { P }{ 2 }$$

C
$$\dfrac { P }{ 4 } $$

D
4P

Solution