Ray Optics and Optical Test

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Ray Optics and Optical Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

A $$60\ watt$$ bulb is hung over the center of a table $$4m \times 4\ m$$ at a height of $$3\ m$$. The ratio of the intensities of illumination at a point on the centre of the edge an don the corner of the table is

A
$$(17/13)^{3/2}$$

B
$$2/1$$

C
$$17/13$$

D
$$5/4$$

Solution

The Illuminance at $$A$$ is
$$I_A=\dfrac{L}{(\sqrt{13})^2} \times \cos \theta_1=\dfrac{L}{13} \times \dfrac{3}{\sqrt{13}} =\dfrac{3L}{(13)^{3/2}}$$
The illuminance at $$B$$ is
$$T_B=\dfrac{L}{(\sqrt{17})^2} \times \cos \theta_2$$
$$=\dfrac{L}{17} \times \dfrac{3}{\sqrt{17}}=\dfrac{3L}{(17)^{3/2}}$$
$$\therefore \dfrac{I_A}{I_B}=\left(\dfrac{17}{13}\right)^{3/2}$$
Question 2
1 / -0

At sun rise of sunset, the sun looks more red than at mid-day because

A
The sun is hottest at these times

B
Of the scattering of light

C
Of the effects of refraction

D
Of the effects of diffraction

Solution

$$I\propto\dfrac{1}{{\lambda}^4}$$
According to Rayleigh’s law of scattering, intensity scattered is inversely proportional to the forth power of wavelength. So red is least scattered and sun appears Red.
Question 3
1 / -0

Refractive index of glass is $$\dfrac {3}{2}$$ and refractive index of water is $$\dfrac {4}{3}$$. If the speed of light in glass is $$2.00\times 10^8\ m/s$$ , the speed in water will be

A
$$2.67\times 10^8 m/s$$

B
$$2.25\times 10^8 m/s$$

C
$$1.78\times 10^8 m/s$$

D
$$1.50\times 10^8 m/s$$

Solution

As we know,
$$\mu=\dfrac{c}{v}$$
$$\mu\propto \dfrac {1}{v}$$ where $$\mu$$
is refractive index of medium , $$c$$ speed of light in vaccum and $$v$$ is speed of light in medium $$\Rightarrow \dfrac {\mu_g}{\mu_w}=\dfrac {v_w}{v_g}\Rightarrow \dfrac {3/2}{4/3}=\dfrac {v_w}{2\times 10^8}$$
$$\Rightarrow v_w =2.25\times 10^8 m/s$$
Question 4
1 / -0

The relation between the linear magnification $$m,$$ the object distance $$u$$ and the focal length $$f$$ is

A
$$m=\dfrac{f-u}{f}$$

B
$$m=\dfrac{f}{f-u}$$

C
$$m=\dfrac{f+u}{f}$$

D
$$m=\dfrac{f}{f+u}$$

Solution

$$\because m=-\dfrac{v}{u}$$ also $$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}\Rightarrow \dfrac{u}{f}=\dfrac{u}{v}+1$$
$$\Rightarrow -\dfrac{u}{v}=1-\dfrac{u}{f}\Rightarrow \dfrac{-v}{u}=\dfrac{f}{f-u}$$ so $$m=\dfrac{f}{f-u}.$$
Question 5
1 / -0

Each quarter of a vessel of depth $$H$$ is filed with liquids of the refractive indices, $$n_1, n_2, n_3$$ and $$n_4$$ from the bottom respectively. The apparent depth of the vessel when looked normally is

A
$$\dfrac {H(n_1+n_2+n_3+n_4)}{4}$$

B
$$\dfrac {H\left(\dfrac {1}{n_1}+\dfrac {1}{n_2}+\dfrac {1}{n_3}+\dfrac {1}{n_4}\right)}{4}$$

C
$$\dfrac {(n_1+n_2+n_3+n_4)}{4H}$$

D
$$\dfrac {H\left(\dfrac {1}{n_1}+\dfrac {1}{n_2}+\dfrac {1}{n_3}+\dfrac {1}{n_4}\right)}{2}$$

Solution

Apparent depth of bottom
$$=\dfrac {H/4}{\mu_1}+\dfrac {H/4}{\mu_2}+\dfrac {H/4}{\mu_3}+\dfrac {H/4}{\mu_4}$$

$$=\dfrac {H}{4}\left(\dfrac {1}{\mu_1}+\dfrac {1}{\mu_2}+\dfrac {1}{\mu_3}+\dfrac {1}{\mu_4}\right)$$
Question 6
1 / -0

A convex lens is used to form real image of an object on a screen. It is observed that even when the positions of the object and that screen are fixed are two position of the lens to form real images. If the heights of the images are $$4\ cm$$ and $$9\ cm$$ respectively the height of the object is

A
$$2.25\ cm$$

B
$$6.00\ cm$$

C
$$6.50\ cm$$

D
$$36.00\ cm$$

Solution

As we know,
$$O=\sqrt{I_1I_2} \Rightarrow O=\sqrt{4 \times 9}=6\ cm$$
Question 7
1 / -0

When a ray of light passes through the second optical medium with a change in the angle, the phenomenon is known as:

A
Reflection of light

B
Absorption of light

C
Refraction of light

D
None of these

Solution

The bending of rays of light after passing through one medium to another is known as refraction.
A light ray traveling obliquely from a denser medium to a rarer medium bends away from the normal. A light ray bends towards the normal when it travels obliquely from a rarer to a denser medium.
Question 8
1 / -0

The length of a simple astronomical telescope is equal to:

A
Difference between focal length of two lenses

B
Half of the sum of focal distances

C
Sum of the focal distances

D
Multiplication of the focal distances

Solution

Length of a simple astronomical telescope
$$L = f_{0} + f_{e}$$
Question 9
1 / -0

The sky would appear red instead of blue if

A
Atmospheric particles scatter blue light more than red light

B
Atmospheric particles scatter all colours equally

C
Atmospheric particles scatter red light more than the blue light

D
The sun was much hotter

Solution

The Colour of the sky is highly scattered light (color). If red color scattered more then the sky would appear red instead of blue.
Question 10
1 / -0

Two point light source are $$24\ cm$$ apart. Where should a convex lens of focal length$$9\ cm$$ be put in between them from one source so that the images of both the sources are formed at the same place.

A
$$6\ cm$$

B
$$9\ cm$$

C
$$12\ cm$$

D
$$15\ cm$$

Solution

The given condition will be satisfied only if one source $$(S_1)$$ placed on one side such that $$u < f$$ ( i.e. it lies under the focus).
The other source $$(S_2)$$ is placed on the other side of the lens such that $$u > f$$ (i.e. it lies beyond the focus).
If $$S_1$$ is the object for lens then $$\dfrac 1f=\dfrac {1}{-y}-\dfrac{1}{-x}$$
$$\Rightarrow \dfrac 1y=\dfrac 1x-\dfrac 1f$$ ......(i)
If $$S_2$$ is the object for lens then
$$\dfrac 1f=\dfrac{1}{+y}-\dfrac{1}{(-24-x)}\Rightarrow \dfrac 1y=\dfrac 1f-\dfrac {1}{(24-x)}$$ ....(ii)
From equation (i) and (ii)
$$\dfrac 1x-\dfrac 1f=\dfrac 1f-\dfrac {1}{(24-x)}\Rightarrow \dfrac 1x+\dfrac {1}{(24-x)}=\dfrac 2f=\dfrac 29$$
$$\Rightarrow x^2-24x+108=0$$. After solving the equation $$x=18\ cm, 6\ cm$$

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Re-Attempt Weekly Quiz Competition

Ray Optics and Optical Test - 60

Result

Ray Optics and Optical Test - 60

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SHARING IS CARING

Question 1

$$(17/13)^{3/2}$$

$$2/1$$

$$17/13$$

$$5/4$$

Solution

Question 2

The sun is hottest at these times

Of the scattering of light

Of the effects of refraction

Of the effects of diffraction

Solution

Question 3

$$2.67\times 10^8 m/s$$

$$2.25\times 10^8 m/s$$

$$1.78\times 10^8 m/s$$

$$1.50\times 10^8 m/s$$

Solution

Question 4

$$m=\dfrac{f-u}{f}$$

$$m=\dfrac{f}{f-u}$$

$$m=\dfrac{f+u}{f}$$

$$m=\dfrac{f}{f+u}$$

Solution

Question 5

$$\dfrac {H(n_1+n_2+n_3+n_4)}{4}$$

$$\dfrac {H\left(\dfrac {1}{n_1}+\dfrac {1}{n_2}+\dfrac {1}{n_3}+\dfrac {1}{n_4}\right)}{4}$$

$$\dfrac {(n_1+n_2+n_3+n_4)}{4H}$$

$$\dfrac {H\left(\dfrac {1}{n_1}+\dfrac {1}{n_2}+\dfrac {1}{n_3}+\dfrac {1}{n_4}\right)}{2}$$

Solution

Question 6

$$2.25\ cm$$

$$6.00\ cm$$

$$6.50\ cm$$

$$36.00\ cm$$

Solution

Question 7

Reflection of light

Absorption of light

Refraction of light

None of these

Solution

Question 8

Difference between focal length of two lenses

Half of the sum of focal distances

Sum of the focal distances

Multiplication of the focal distances

Solution

Question 9

Atmospheric particles scatter blue light more than red light

Atmospheric particles scatter all colours equally

Atmospheric particles scatter red light more than the blue light

The sun was much hotter

Solution

Question 10

$$6\ cm$$

$$9\ cm$$

$$12\ cm$$

$$15\ cm$$

Solution

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