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CBSE Questions for Class 10 Physics Human Eye And Colorful World Quiz 14 - MCQExams.com
CBSE
Class 10 Physics
Human Eye And Colorful World
Quiz 14
A ray falls on a prism ABC (AB = BC) and travels as shown in figure. The minimum refractive index of the prism material should be
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$$\dfrac{4}{3}$$
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$$\sqrt{2}$$
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$$1.5$$
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$$\sqrt{3}$$
Explanation
$$\begin{array}{l} By\, the\, help\, of\, the\, figure \\ { \sin ^{ -1 } }\left( { \frac { { nr } }{ { nd } } } \right) =c \\ \Rightarrow \frac { { nr } }{ { nd } } =\sin { 45^{ \circ } } \\ \Rightarrow \frac { 1 }{ { nd } } =\frac { 1 }{ { \sqrt { 2 } } } \\ \therefore nd=\sqrt { 2 } \\ Hence,\, the\, option\, B\, \, is\, \, the\, correct\, answer. \end{array}$$
A presbyopic patient has near point as $$30\ cm$$ and far point as $$40\ cm$$. The dioptric power for the corrective lens for seeing distant objects is
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$$40\ D$$
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$$4\ D$$
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$$-2.5\ D$$
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$$0.25\ D$$
Explanation
In this case, for seeing distant objects the far point is $$40\ cm$$.
Hence the required focal length is
$$f=-d$$ ( distance of a far point ) $$=-40\ cm$$
Power $$P=\dfrac{100}{f}\ cm=\dfrac{100}{-40}=-2.5\ D$$
The eyes defect represented by the figure is
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Myopia
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Hypermetropia
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cataract
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presbyopia
A man can see clearly up to $$3\ \textit{metres}$$. Prescribe a lens for his spectacles so that he can see clearly up to $$12\ \textit{metres}$$
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$$-3/4\ D$$
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$$3\ D$$
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$$-1/4\ D$$
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$$-4\ D$$
Explanation
For corrective lens to be used , object distance takes as desired distance person wants to see , so $$u=$$ wants to see $$=-12\ m$$
and image distance taken as distance person initially able to see , so $$v=$$can see $$=-3\ m$$
$$\therefore P=\dfrac 1f=\dfrac 1v-\dfrac 1u =\dfrac{1}{-3}-\dfrac{1}{(-12)}=-\dfrac 14\ D$$
A person cannot see distinctly at the distance less than one metre. Calculate the power of the lens that he should use to read a book at a distance of $$25\, cm$$
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$$+ 3.0\, D$$
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$$+ 0.125\, D$$
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$$- 3.0\, D$$
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$$+ 4.0\, D$$
Explanation
As we know,
$$P=\dfrac 1f=-\dfrac 1v+\dfrac 1u=-\dfrac{1}{100}+\dfrac{1}{25}=\dfrac{3}{100}=+3\ D$$
A rainbow is formed because water droplets of the atmosphere behave like glass slabs.
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True
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False
Explanation
A rainbow is formed because water droplets behave like prism.
Hence, the statement is false.
In a glass prism, spectrum is produced due to :
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refraction
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dispersion
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scattering
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diffraction
Explanation
The refractive index of many materials (such as
glass
) varies with the wavelength or color of the light used. This causes light of different colors to be refracted differently and to leave the
prism
at different angles. this
phenomenon known as dispersion.
The speeds of red light and yellow light are exactly same
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in vacuum but not in air
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in air but not in vacuum
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in vacuum as well as in air
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neither in vacuum nor in air.
A normal eye is not able to see objects closer than $$25 \ cm$$ because?
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The focal length of the eye is $$25 \ cm$$
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The distance of the retina from the eye-lens is $$25 \ cm$$
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The eye is not able to decrease the distance between the eye-lens and the retina beyond a limit
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The eye is not able to decrease the focal length beyond a limit
Explanation
For a ‘normal eye’ to visualize the objects it must have a ‘focal length’ of greater than $$25 \ cm$$. If the eye has to focus particles ‘less than $$25 \ cm$$’ then the ciliary muscles of the eyes should expand such as to change its ‘focal length’.
The inability of the eye to decrease the focal length beyond a limit is the reason the eye is not able to see objects closer than $$25 \ cm$$.
Hence option D is correct
One can not see through fog because:
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fog absorbed light
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light is scattered by the droplets in fog
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light surfers total reflection by the droplets in the fog
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the refractive index of fog is in infinity
Explanation
The fog certains water droplets, When the reflected light from an object travels through fog, it interacts with the water droplets of the fog and is scattered by them. The scattering of light by fog droplets is not uniform in all directions. also it doesn't reaches to the observer eyes so we can not cannot see through fog.
The rising and setting of sun appear red because of :
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refraction
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reflection
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defraction
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scattering
Explanation
The reddish appearance of the
sun
at
sunrise
or
sunset
is due to scattering of light by the molecules of air and other fine particles in the atmosphere.
Hence the option D is the right answer
For a person suffering from myopia, the image of the object is formed in front of the retina.
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True
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False
Explanation
In myopia, a person is not able to see far away objects due to the elongation of the eyeball. The image is formed in front of the retina as shown in the figure above.
Abnormalities in the normal vision of the eye are called defects of vision.
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True
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False
Explanation
When the person is not able to see the objects clearly or he is having difficulty in seeing the objects he may have defects of vision like myopia, hypermetropia, presbyopia, astigmatism.
Short-sightedness is caused due to elongation of the eyeball.
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True
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False
Explanation
Myopia is also known as near-sightedness. A person with myopia can see nearby objects clearly but cannot see distant objects distinctly.
This defect may arise due to:
$$i)$$ excessive curvature of the eye lens
$$ii)$$ elongation of the eyeball
Write true or false for the following statements :
Presbyopia is a type of far-sightedness.
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True
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False
Explanation
Presbyopia is an age-related condition in which the lens of the eye becomes less flexible. Seeing details like words in a book or an online article, or adjusting focus between far away and nearby objects is difficult this condition is most common in people between the ages of $$ 40 $$ and $$ 50 $$.
While in far-sightedness it's hard to see things close up. People of any age, including babies, can be farsighted.
Write true or false for the following statements :
A near sighted person can see only nearer objects clearly.
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True
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False
Explanation
Nearsightedness is a vision condition in which one can see nearer objects clearly, but objects farther away appear blurry. It occurs when light rays are focused in front of the retina instead of on retina.
Hence, the given statement that a near sighted person can see only nearer objects clearly is True.
A rainbow can be seen in the sky
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When the sun is in front of you.
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When the sun is behind you.
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When the sun is overhead.
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Only at the time of sun rise.
Explanation
The correct answer is option (b). when the sun is behind you. You can easily see rainbows in the rainy season when the sun is behind you and the sunlight is low.
The eye defect represented by the figure is
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Myopia
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Hypermetropia
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Cataract
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Presbyopia
Explanation
In the given figure, Diverging lens are used for clear viewing. Its means that defected eyes was making image of an object inform of ratina. This defect of eyes is known as Myopia.
Option A
The color of the sky is blue during the daytime, red during sunset, and black at night. This is due to
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Scattering of light
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Small particles present in the atmosphere
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Atmospheric refraction
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All of the above.
Explanation
A and C. Gases and particles in Earth's atmosphere scatter sunlight in all directions. Blue light is scattered more than other colors because it travels as shorter, smaller waves. This is why we see a blue sky most of the time.
B. During sunrise and sunset the sun is low in the sky, and it transmits light through the thickest part of the atmosphere. We see the red because red wavelengths (the longest in the color spectrum) are breaking through the atmosphere. The shorter wavelengths, such as blue, are scattered and broken up.
A student of class $$ 10, $$ is not able to see clearly the blackboard question when seated at a distance of $$ 5\ \mathrm{m} $$
from the board, the defect he is suffering from is
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Myopia
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Hypermetropia
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Presbyopia
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Astigmatism
Explanation
Myopia is a vision condition in which you can see nearby objects clearly but far away objects appear blurry. In this defect, the light from far away objects focusses in front of the retina.
Since, the student is not able to see blackboard, which is located at a far away($$5 \ m$$) distance, clearly. Hence, he is suffering from Myopia.
Long-sightedness or hypermetropia can be corrected by
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Planar lens
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Concave lens
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Convex lens
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Bifocal lens
Explanation
Convex lens
The colored light that refracts most while passing through a prism is
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Yellow
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Violet
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Blue
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Red
Explanation
The white light is dispersed into its seven-color components by a prism. Different colors of light bend through different angles with respect to the incident ray, as they pass through a prism. The red light bends the least while the
violet the most.
A man can see the object up to a distance of one meter from his eyes. For correcting his eyesight so that he can see an object at infinity, he requires a lens whose power is
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$$+0.5\ D$$
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$$+1.0\ D$$
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$$+2.0\ D$$
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$$-1.0\ D$$
Explanation
The image of object at infinity should be formed at $$100\ cm$$ from the eye
Apply formula:
$$\dfrac 1f =\dfrac{1}{v}-\dfrac{1}{u}$$
$$f$$ : focal length
$$u$$ : object distance
$$v$$ : Image distance
$$\dfrac 1f =\dfrac{1}{\infty}-\dfrac{1}{100}$$
$$\Rightarrow \dfrac{1}{f}=-\dfrac{1}{100}$$
Apply formula of power:
$$P = \dfrac{1}{f}$$
So the power $$=\dfrac{-100}{100}=-1\ D$$
( Distance is given in $$cm$$ but $$P=\dfrac 1f$$ in metres )
A man can see the object between $$15\ cm$$ and $$30\ cm$$. He uses the lens to see the far objects. Then due to the lens used, the near point will be at
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$$\dfrac{10}{3}\ cm$$
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$$30\ cm$$
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$$15\ cm$$
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$$\dfrac{100}{3}\ cm$$
Explanation
For improving far point, concave lens is required and for this concave lens $$u=\infty, v=-30\ cm$$
So,
$$\dfrac 1f=\dfrac{1}{-30}-\dfrac{1}{\infty}$$
$$\Rightarrow f=-30\ cm$$
for near point,
$$\dfrac{1}{-30}=\dfrac{1}{-15}-\dfrac 1u $$
$$\Rightarrow u=-30\ cm$$
A man suffering from myopia can read a book placed at $$10\ cm$$ distance. For reading the book at a distance of $$60\ cm$$ with relaxed vision, the focal length of the lens required will be:
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$$45\ cm$$
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$$-20\ cm$$
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$$-12\ cm$$
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$$30\ cm$$
Explanation
According to the question,
Far point of the myopic eye $$= 10\ cm$$
Far point of a normal eye $$= 60\ cm$$
object distance, $$u = -60\ cm$$
image distance, $$v = -10\ cm$$
focal length, $$f$$
Using lens formula
$$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$$
$$\dfrac{1}{f} = \dfrac{1}{-10} - \dfrac{1}{\left(-60\right)} = \dfrac{-6 + 1}{60}$$
$$\dfrac{1}{f} = \dfrac{-5}{60}$$
$$f = -12\ cm$$
The far point of a myopia eye is at $$40\ cm$$. For removing this defect, the power of lens required will be
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$$40\ D$$
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$$-4\ D$$
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$$-2.5\ D$$
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$$2.5\ D$$
Explanation
For myopic eye $$f=-$$ ( defect far point )
$$\Rightarrow f=-40\ cm\Rightarrow P=\dfrac{100}{-40}=-2.5\ D$$
A person is suffering from myopic defect. He is able to see clear objects placed at $$15\ cm$$. What type and of what focal length of lens he should use to see clearly the objects placed $$60\ cm$$ away
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Concave lens of $$20\ cm$$ focal length
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Convex lens of $$20\ cm$$ focal length
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Concave lens of $$12\ cm$$ focal length
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Convex lens of $$12\ cm$$ focal length
Explanation
Lens should be used such that the image of an object placed at $$60 \ cm$$ is formed at $$15 \ cm$$
Hence, in the given case $$u=-60 \ cm$$; $$v=-15 \ cm$$
Using the lens' formula $$\dfrac 1f=\dfrac{1}{v}-\dfrac{1}{u}$$
$$\Rightarrow \dfrac 1f=\dfrac{1}{-15}-\dfrac{1}{-60} = \dfrac{-3}{60}$$
$$\Rightarrow f=-20\ cm$$.
Hence, the person should use concave lens of focal length $$20 \ cm$$
The hyper-metropia is a
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Short-side defect
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Long-side defect
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Bad vision due to old age
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None of these
Explanation
The hyper-metropia is an eye defect when person faces difficulty in seeing nearby objects(Short Sight) but not far-sided objects (Long Sight). Due to inability of the eye lens to converge light rays from nearby objects at the retina, image is formed behind the retina. Long sight leads to problems with near vision. So hyper-metropia is a Long-side defect.
When white light passes through the achromatic combination of prisms, then what is observed
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Only deviation
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Only dispersion
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Deviation and dispersion
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None of the above
Explanation
When white light passes through a prism , it disperses into band of seven colours. But when two prism are combined in such a way that sum of angular dispersions of crown glass prism and flint glass prism is zero then such a combination is achromatic combination of prisms. When white light passes through achromatic combination of prisms, internally dispersed components of white light from Crown glass prism refract and meet together at the outer surface edge of Flint Glass prism ,so the refracted light from the achromatic combination of prisms become white light again with deviation only without any dispersion.
Stars are twinkling due to
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Diffraction
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Reflection
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Refraction
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Scattering
Explanation
Stars twinkle due to variation in $$R.l$$ of atmosphere.
Stars are twinkling due to refraction,
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