MCQ Questions for Class 10 Physics Human Eye And Colorful World Quiz 13

A long-sighted person cannot see objects clearly at a distance less than

$40\ cm$ from his eye. The power of the lens needed to read an object at

$25\ cm$ is

0%
$-2.5 D$
0%
$+2.5 D$
0%
$-6.25 D$
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$+1.5 D.$

Explanation

$\textbf{Hint}$ Use the lens formula.

The sky appears blue, due

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To scattering of light by dust particles
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To the colour of the sea water
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To the presence of water in clouds
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None of these

Which of the following statements is correct about rainbow?

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In primary rainbow, red colour on the outside and violet colour on the inside.
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In primary rainbow, violet colour on the outside and red colour on the inside.
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Secondary rainbow is brighter than primary rainbow.
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In secondary rainbow, light waves suffers one total internal reflection before coming out.

The nearer point of hypermetropic eye is 40 cm. The lens to be used for its correction should have the power?

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+ 1.5 D
0%
- 1.5 D
0%
+ 2.5 D
0%
+ 0.5 D

Explanation

Near point of a hypermetropic eye

$= 40\ cm$

Near point of a normal eye

$= 25\ cm$

Object distance,

$u = -25\ cm$

Image distance,

$v = -40\ cm$

Using lens formula,

$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{-40} + \dfrac{1}{25}$

$\dfrac{1}{f} = \dfrac{-5 + 8}{200} = \dfrac{3}{200\ cm} = \dfrac{300}{200\ m}$

$P = \dfrac{1}{f {\text{(in m)}}} = \dfrac{300}{200} = \dfrac{3}{2}$

$P = 1.5\ D$

In the given figure, if P is a point of the source of light, R is retina and L is the lens of the eye, the person having such condition of his eyes is suffering from :

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Hypermetropia
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Myopia
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Cataract
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Night Blindness

Which of the following statements is correct concerning the passage of white light into a glass prism?

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All the colours of white light travel with the same speed
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The violet colour travels slower than the red colour
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The violet colour travels faster than the red colour
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Greater the wavelength, slower the speed of colour

Of the following natural phenomena, tell which one is based on the refraction of light:

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Earthlight
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Polar light
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Rainbow
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Lightning

A ray of light containing both red and blue colours is incident on the refracting surface of a prism. Then,

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both colours suffer equal deviation
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red colour suffers more deviation
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red colour suffers less deviation
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deviation depends on the angle of prism

A person cannot see object clearly that are closed that

$2m$ and father than

$4m$ . To correct the eye vision the person will use :

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Bifocal lenses of power $0.5D$ and $0.25D$
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Bifocal lenses of power $0.25 D$ and $3.5D$
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Bifocal lenses of power $0.5 D$ and $4.0 D$
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Bifocal lenses of power $4.0 D$ and $0.5 D$

Explanation

Given,

The distance of the far point

$(D) = 2m$

Let f be the focal length of the eye lens.

Image distance ( distance between eye lens and retina ) for our human eye is always constant and is about

$0.25m.$

This is due to the action of ciliary muscles.

applying lens formula, we get

$f = -D = -2 m.$

But, the power of a lens is reciprocal of focal length.

Hence. Power

$=\dfrac{1}{f}.$ ( f in meters )

$\dfrac{1}{-2}=0.5$ D

Similarly,

The distance of the far point

$(D) = 4m$

Let f be the focal length of the eye lens.

Image distance ( distance between eye lens and retina ) for our human eye is always constant and is about

$0.25m.$

This is due to the action of ciliary muscles.

applying lens formula, we get

$f = -D = -4 m.$

But, the power of a lens is reciprocal of focal length.

Hence. Power

$=\dfrac{1}{f}.$ ( f in meters )

$\dfrac{1}{-4}=0.25$ D

So he has to use by focal lenses

A man wearing glasses of focal length + 1m cannot clearly see beyond 1 m

0%
if he is farsighted
0%
if he is nearsighted
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if his vision is normal
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in each of these cases.

To remove myopia (short-sightedness) a lens of power

$- 0.66D$ is required. The distant point of the eye is approximate.

0%
$100\ cm$
0%
$151.5\ cm$
0%
$50\ cm$
0%
$25\ cm$

Explanation

The given power

$(P)$ is

$-0.66D$ .

We know that-

$P = \dfrac{ 1 }{ f }$

$f$ ; focal length

$\Rightarrow f = \dfrac{1}{P}$

$\Rightarrow f = \dfrac{ 1 }{ -0.66 }$

$\Rightarrow f = 151.5\ cm$

Hence, the distant point of the eye is

$151.5\ cm$ .

A man wearing glasses of power

$+2D$ , can read clearly a book placed at a distance of

$40\ cm$ , from the eye. The power of the lens required, so that he can read at

$25\ cm$ from the eye is ?

0%
$+4.5\ D$
0%
$+4.0\ D$
0%
$+3.5\ D$
0%
$+3.0\ D$

Explanation

$P = 2$

$\Rightarrow f = \dfrac{1}{P} = \dfrac{1}{2}\ m = 50\ cm$

$u=-40cm$

$\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

$\Rightarrow \dfrac{1}{50} = \dfrac{1}{v} - \dfrac{1}{-40}$

$\Rightarrow \dfrac{1}{50} - \dfrac{1}{40} = \dfrac{1}{v}$

$\Rightarrow v = -200\ cm$

Now for new lens object distance = new near point =-

$25cm$

$u_{2} = -25\ cm$

$v_{2} = -200\ cm$

$\Rightarrow \dfrac{1}{f} = \dfrac{1}{v_2} - \dfrac{1}{u_2}$

$= \dfrac{1}{-200} - \dfrac{1}{-25}$

$\dfrac{1}{f} = \dfrac{1}{25} - \dfrac{1}{200} = \dfrac{7}{200}\ cm^{-1} = \dfrac{7}{2}\ m^{-1}$

$P =\frac{1}{f}$ =

$+3.5\ D$

In the figure below for no dispersion to take place, light rays must enter from _____ and emerge from _____.

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$A_1 B_1, \ P_1 R_1$
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$A_2 B_2, \ P_2 Q_2$
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$A_1 B_1, \ Q_1 R_1$
0%
$A_2 B_2, \ R_2 Q_2$

Explanation

The light ray must enter from

$A2B2$ and emerge from

$P2Q2$

A person cannot see the object beyond 200cm. The power of lens used to correct the vision will be :

0%
+0.5 D
0%
+5D
0%
-0.5D
0%
-5D

Explanation

Here for the lens needed , it should form image of an distant object. at distance of

$200 cm$

Here,

$u = \infty$ and

$v = -200\; cm$

$\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

$\Rightarrow \dfrac{1}{f} =- \dfrac{1}{200} + 0$

$\Rightarrow f =-2m$

$\Rightarrow P (in \; D)= \dfrac{1}{f(in\; m)} =- 0.5 D$

Therefore, C is correct option.

An old person is able to see an object nearest to 1 m.What should be the power of lens required so that he can see an object placed at nearest distance of distinct vision?

0%
3D
0%
4D
0%
1/2D
0%
1/4D

Explanation

Hypermetropia is a condition of the eye where the person is not able to see things clearly when nearer to the eye.

The normal near point of the eye is = $25cm$

For corrective lens to be used ,

Focal length,= $f$

For corrective lens object distance =The normal near point of the eye = $u=-25cm$ and image distance = defected near point of man = $v=-100cm$

Lens formula:

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

$\dfrac{1}{f}=\dfrac{1}{-100}-\dfrac{1}{-25}=\dfrac{1}{25}-\dfrac{1}{100}=\dfrac{3}{100}$

$f=33.33\,cm=0.33\,m$

Power, $P=\dfrac{1}{f}=\,\dfrac{1}{0.333}=3D$

Hence, Power of lens is $3D$

A far-sighted person cannot focus distinctly objects closer than 120 cm. The lens that will permit him to read from a distance of 40 cm will have a focal length:

0%
+ 30 cm
0%
- 30 cm
0%
+ 60 cm
0%
- 60 cm

Explanation

Given,

Near point, $v=-120\,cm$

Reading Distance, $u=-40\,cm$

From lens formula,

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{-120}-\dfrac{1}{-40}=\dfrac{1}{60}$

$f=+60\,cm$

Focal length of lens $f=+60\,cm$

A person can not see the distant objects clearly (though he can see the nearby object clearly). He is suffering from the defect of vision called:

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cataract
0%
hypermetropia
0%
myopia
0%
presbyopia

A person cannot see distant objects clearly. his vision can be corrected by using the spectacle containing :

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concave lenses
0%
plane lenses
0%
contact lenses
0%
convex lenses

Rainbow is observed when the Sun is

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In any of these position
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In front of the observer
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Behind the observer
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Vertically above the observer

A person can see clearly objects only when they lie between

$50$ cm and

$400$ cm from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens, the person has to use, will be?

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Convex, $+2.25$ dioptre
0%
Concave, $-0.25$ dioptre
0%
Concave, $-0.2$ dioptre
0%
Convex, $+0.15$ dioptre

Explanation

Maximum distance of distinct vision

$=400$ cm. So image of object at infinity is to be formed at

$400$ cm

Use lens formula,

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

$\dfrac{1}{-400}-\dfrac{1}{\infty}=\dfrac{1}{f}$

$P=-0.25$ D.

A young man has to hold a book at arm's length to be able to read it clearly. The defect of vision is:

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astigmatism
0%
myopia
0%
presbyopia
0%
hypermetropia

Though a woman can see the distant objects clearly, she cannot see the nearby object clearly. She is suffering from the defect called :

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Long-sight
0%
Short-sight
0%
Hind-sight
0%
Mid-sight

What is the common phenomena of light between random wavering of flickering of objects seen through a turbulent stream of hot air, advance sunrise and delayed sunset and twinkling of stars?

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Refraction
0%
Reflection
0%
Scattering
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Atmospheric refraction

The near point and the far point of a child are at 10 and 100 cm. If the retina is 2.0 cm behind the eyelens .What is the range of the power of the eye lens?

0%
$50 D \text { to } 40 \mathrm { D }$
0%
$60 D \text { to } 51 \mathrm { D }$
0%
$60 D \text { to } 54 \mathrm { D }$
0%
$40 D \text { to } 50 \mathrm { D }$

After looking the sun the colour of other things appears

0%
Black
0%
White
0%
Yellow
0%
Red

A person suffering from short-sightedness can see clearly only upto a distance of 2 metres. Find the power of the lens required to correct his vision

0%
-0.5 D
0%
-1.5 D
0%
+2.5 D
0%
+1.5 D

Explanation

To correct the vision, the object is placed at infinity

$u=-\infty$ so that its image is formed at his far point, i.e.

$v=-2 \ m$

$\begin{array}{l} { { Re } }quired\, lens\, is\, a\, concave\, lens. \\ v=-2 \ m \\ u=\infty \\ Now, using \ Lens \ Formula \\ \dfrac { 1 }{ v } -\dfrac { 1 }{ u } =\dfrac { 1 }{ f } \\ \dfrac { 1 }{ { -2 } } -\dfrac { 1 }{- \infty } =\dfrac { 1 }{ f } \\ f=-2m \\ Power \ P=\dfrac { 1 }{ f } =\dfrac { 1 }{- 2 } =-0.5D \end{array}$

Hence, the option

$A$ is the correct answer.

The phenomenon of splitting of sunlight by a prism is called_________ and the color that deviates most is ______

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Reflection,violet
0%
Refraction,Red
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Scattering,Yellow
0%
Dispersion,violet

Hypermetropia is possible in.........

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children
0%
aged person's
0%
high intensity fall on eye
0%
low intensity fall on eye

A student of class

$10,$ is not able to see clearly the black board question when seated at a distance of

$5 \mathrm{m}$ from the board,the defect he is suffering from is

0%
Myopia
0%
Hypermetropia
0%
Presbyopia
0%
Astigmatism

The principle section of a glass prism is isosceles triangle ABC with AB=AC, the face AC is silvered. A ray of light is incident normally on the face AB and after two reflections, it emerges from the base BC perpendicular to the base. Angle of BAC of the prism is

0%
30 $^{ \circ }$
0%
36 $^{ \circ }$
0%
18 $^{ \circ }$
0%
72 $^{ \circ }$

CBSE Questions for Class 10 Physics Human Eye And Colorful World Quiz 13 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

CBSE Questions for Class 10 Physics Human Eye And Colorful World Quiz 13 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

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