Match the column I with column II and select the correct option from the codes given below. Column I Column II (a) Choroid (i) Detects light stimulus (b) Retina (ii) Absorbs light and prevent it from being reflected within the eyeball (c) Cornea (iii) Controls the size of the pupil (d) Iris (iv) Helps to focus light as it enters the eye.

(a)-(ii). (b)-(i), (c)-(iv), (d)-(iii)

(a)-(i), (b)-(ii), (c)-(iii). (d)-(iv)

(a)-(ii). (b)-(iii), (c)-(iv), (d)-(i)

(a)-(iii), (b)-(iv), (c)-(i), (d)-(iii)

MCQ Questions for Class 10 Physics Human Eye And Colorful World Quiz 16

Which of the following phenomena cannot be explained by scattering of light?

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Blue colour of sky
0%
White colour of clouds
0%
Tyndall effect
0%
Formation of rainbow

A person can see clearly objects between 15 and 100 cm from his eye. The range of his vision if he wears close fittings spectacles having a power of

$0.8$ dipole is

0%
$5$ to $500cm$
0%
$12$ to $250cm$
0%
$17$ to $500cm$
0%
$17$ to $250cm$

Read the given statements and mark the correct option.
Statement 1 : Abeam of white light gives a spectrum on passing through a hollow prism.
Statement 2 : Speed of light outside the hollow prism is different from the speed of light inside the hollow prism.

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Both statements 1 and 2 are true and statement 2 is the correct explanation of statement 1.
0%
Both statements 1 and 2 are true but statement 2 is not the correct explanation of statement 1.
0%
Statement 1 is true but statement 2 is false.
0%
Both statements 1 and 2 are false.

A large-angle prism deviates a light ray

0%
more
0%
less
0%
equal to a small-angle prism
0%
does not deviate at all

Explanation

Let,

$i=\text{angle of incidence}$ ,

$e=\text{angle of emergence}$ and

$A=\text{angle of prism}$

$\begin{array}{l} \text { We know for prism, deviation } \\ \qquad \delta=i+e-A \\ \text { If } A \text { is large } \delta \text { is less by this equation } \\ \text { so option (B) less is the right option } \end{array}$

2006013_1015160_ans_cc3dee01ad9b44a5926a04d221b7bc4a.png

Ocean appears blue due to _____________________.

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Greater depth
0%
Scattering of light by water particles
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Blue colour of water
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None of these

Which defect of vision of eye is shown respectively?

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P: Presbyopia, Q: Near sightedness
0%
P: Far sightedness, Q: Near sightedness
0%
P: Near sightedness, Q: Far sightedness
0%
P: Presbyopia, Q: Far sightedness

Explanation

$P:-$

In this we see that the eye can't see the nearby objects as the lens can't focus the image on the retina but focuses the image beyond the retina.

This defect of the eye is called

$far-sightedness$ as the eye in this defect can't see the nearby objects but can see far-off objects.

$Q:-$

In this we see that the eye can't see the far off objects clearly as the lens can't focus the image on the retina but focuses the image before the retina.

This defect of the eye is called

$near-sightedness$ as the eye in this defect can't see the far away objects but can see nearbyf objects.

Answer-(B)

In the experiment of prism incident red color ray instead of white light through a prism. Which color band will you observe on screen?

0%
Red colour spectrum
0%
All the colours of rainbow
0%
Maroon and yellow colour spectrum
0%
White colour spectrum

Match the column I with column II and select the correct option from the codes given below.

	Column I		Column II
(a)	Choroid	(i)	Detects light stimulus
(b)	Retina	(ii)	Absorbs light and prevent it from being reflected within the eyeball
(c)	Cornea	(iii)	Controls the size of the pupil
(d)	Iris	(iv)	Helps to focus light as it enters the eye.

0%
(a)-(ii). (b)-(i), (c)-(iv), (d)-(iii)
0%
(a)-(i), (b)-(ii), (c)-(iii). (d)-(iv)
0%
(a)-(ii). (b)-(iii), (c)-(iv), (d)-(i)
0%
(a)-(iii), (b)-(iv), (c)-(i), (d)-(iii)

The near point of a person is

$75cm$ . In order that he may be able to read book at a distance

$25cm$ . The power of spectacles lenses should be

0%
$-2D$
0%
$+3.75D$
0%
$+2.6D$
0%
$+3D$

A far sighted man cannot focus distinctly on objects closer than

$1m$ . What is the power of the lens that will permit him to read from a distance of

$40 \,cm$ .

0%
$2.5 D$
0%
$-2.5 D$
0%
$1.5 D$
0%
$1.67 D$

Which phenomenon is responsible for the twinkling of stars?

0%
Atmospheric reflection
0%
Atmospheric refraction
0%
Reflection
0%
Total internal reflection

For a normal eye, the cornea of eye provides as converging power of 40 D and the lens least converging power of the eye lens behind the cornea is 20 D. Using this information , the distance between the retina and the cornea – eye lens can be estimated to be:

0%
1.5 cm
0%
5 cm
0%
2.5 cm
0%
1.67 cm

Explanation

Given that,

Power ${{P}_{1}}=40\,D$

Power ${{P}_{2}}=20\,D$

Total power of the combination

$P={{P}_{1}}+{{P}_{2}}$

$P=40+20$

$P=60\,D$

Now, focal length of the combination

$f=\dfrac{1}{60}\times 100$

$f=\dfrac{5}{3}\,cm$

Now, for minimum converging state of the eye lens object is at infinity

$u=-\infty$

$f=\dfrac{5}{3}$

Now, from lens equation

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

$\dfrac{3}{5}=\dfrac{1}{v}-\dfrac{1}{\infty }$

$\dfrac{1}{v}=\dfrac{3}{5}$

$v=\dfrac{5}{3}\,cm$

So, the distance between retina and cornea – lens system $1.67\ cm$

The near point of a hypermetropic person is 75 cm from the eye. What is the power of the lens required to enable the person to read clearly a book held at 25 cm from the eye?

0%
$1\ D$
0%
$1.5\ D$
0%
$0.75\ D$
0%
$2.67\ D$

What does the dispersion of light in a medium imply
i) Lights of different wavelength travel with different speeds in the medium.
ii) Light of any frequency will travel with the same speed in the medium
iii) The refractive index of the medium is different for different wavelength of light

0%
$(i)$ and $(ii)$ only
0%
$(ii)$ and $(iii)$ only
0%
$(i)$ and $(iii)$ only
0%
$(i),(ii)$ and $(iii)$

For an equilateral prism if :

$\angle { i } = angle\ of\ Incidence$

$\angle { e } = angle\ of\ Emergence$

$\angle { P } = angle\ of\ Prism$

$\angle { D } = angle\ of\ Deviation$
then which one relation is correct :

0%
$\angle i + \angle e$ = $\dfrac {\angle P + \angle D}{2}$
0%
$\angle i + \angle D = \angle P + \angle e$
0%
$\angle i + \angle e = \angle P + \angle D$
0%
$None\ of these$

Explanation

$\begin{array}{l} \text { we know, } \\ \qquad i+e=A+\delta \\ i \rightarrow \text { angle of incidence } \\ e \rightarrow \text { angle of emergence } \\ A \rightarrow \text { angle of prism } \\ \delta \rightarrow \text { angle of deviation } \\ \Rightarrow \quad \angle i+\angle e=\angle P+\angle D \end{array}$

After testing the eyes of a child, the optician has prescribed the following lenses for his spectacles:
left eye :

$+ 2\, D$ Right eye:

$+ 2.25\, D$
the child is suffering from the defective vision called:

0%
Short-sightedness
0%
Long-sightedness
0%
Cataract
0%
Presbyopia

Tyndall effect is applicable when?

0%
The diameter of the dispersed particle is not much smaller than the wavelength of the light used
0%
The diameter of the dispersed particles is much smaller than the wavelength of the light used
0%
The refractive indices of the dispersed phase and the dispersion medium must be same
0%
The refractive indices of the dispersed phase and the dispersion medium must differ greatly in magnitude

Explanation

$\begin{array}{l} \text { Tyndall effect is observed only when the following two } \\ \text { conditions are satisfied. } \\ \text { i) The diameter of the dispersed particles is not much } \\ \text { smaller then the wavelength of the light used } \\ \text { ii) The refractive indices of the dispersed phase and the } \\ \text { dispersion medium differ greatly in magnitude. } \\ \text { Hence option (A) and (D) are correct } \end{array}$

A person suffering from an eyesight defect has a far point at

$40\ cm$ and a near the point at

$25\ cm$ . The person uses a lens to see far away object. Find the near point of the person while wearing this lens.

0%
$50\ cm$
0%
$\dfrac{200}{13}\ cm$
0%
$\dfrac{200}{3}\ cm$
0%
$25\ cm$

Explanation

Far point

$=40 \mathrm{~cm}$

$\Rightarrow$ If object is at

$\infty$ , it image should form at

$40 \mathrm{~cm}$ by lens.

Now,

$\begin{aligned}& \frac{1}{40}-0=\frac{1}{f}\\\Rightarrow & f=40\mathrm{~cm}.\end{aligned}$

Near hoint

$=25 \mathrm{~cm}$

With lenses on, the image will be formed at:

$\begin{aligned}\frac{1}{v} &=\frac{1}{40}-\frac{1}{25} \\&=\frac{-3}{200} \\v &=\frac{-200}{3}\mathrm{~cm} .\end{aligned}$

$\text {Ans: } \dfrac{200}{3} \mathrm{~cm}$

Aperson suffering from defective vision can see objects clearly only beyond 100 cm from the eye. Find the power of lens required so that he can see clearly the object placed at a distance of distinct vision (D=25 cm)

0%
2D
0%
3D
0%
-2D
0%
-3D

The intensity of scattered light

0%
$I\propto { \lambda }^{ 4 }$
0%
$I\propto { \lambda }^{ -4 }$
0%
$I\propto { \lambda }^{ -2 }$
0%
$I\propto { \lambda }^{ 2 }$

A far sighted person can see object beyond 71 cm clearly if separation between glasses and eye lens is 2 cm, then find focal length of glass?

0%
23 cm
0%
34.5 cm
0%
18.4 cm
0%
30 cm

Explanation

Complete step by step answer:

We have given that the person can see beyond 71 cm clearly that means for the object placed at distance of distinct vision (25 cm), the image should be formed at 71 cm from the eye. We have given the separation between glasses and eye lens is d=2cm.

Therefore, the object distance from the lens is,

$u=−(25−2)=−23cm$

And the image distance is,

$v=−(71−2)=−69cm$

We have the lens formula relating focal length, image distance and object distance,

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

Substituting

$u=−23cm$

and

$v=−69cm$

in the above equation, we get,

$\dfrac{1}{f}=\dfrac{-1}{69}+\dfrac{1}{23}$

⇒

$f=\dfrac{69}{2}$

$∴f=34.5cm$

Since the focal length is positive, the lens that the person is using is the convex lens.

So, the correct answer is option B.

Note:Do not take the image distance as 71 cm since it is the distance between the eye lens and the image. The image distance and object distance should be taken from the pole of the lens. The distance of distinct vision is 25 cm and it is always negative.

What kind of lens is used in the spectacles of a person suffering from Hypermetropia?

0%
Convex.
0%
Concave.
0%
Plano -convex
0%
A combination of cocave and convex.

In which order , the colour spectrum VIBGYOR appeared to him in the rainbow?

0%
top to bottom
0%
bottom to top
0%
left to right
0%
right to left

Which of the following statements is correct regarding the propagation of light of different colors of white light in the air?

0%
Red light moves fastest
0%
Blue light moves faster than green light
0%
All the colours of the white light move with the same speed in empty space
0%
Yellow light moves with the mean speed as that of the red and the violet light

Explanation

In vacuum, all the colours of white light move with the same speed. They have different wavelengths and frequencies but move with a contant speed

$c = 3 \times 10^8 \ m/s$

To focus the image of a nearby object on the retina of an eye

0%
The distance between eye-lens and retina is increased
0%
The distance of the eye-lens and retina is decreased
0%
The thickness of eye-lens is decreased
0%
The thickness of eye-lens is. increased

Select the correct option and write the completes statements.
What is the reason for twinkling of stars?

0%
Explosions accuring in stars from time to time
0%
Absorption of light in the earth's armosphere.
0%
Motion of stars.
0%
Changing refractive index of the atmospheric gases.

A person can see clearly between 1 m and 2 m his corrective lenses should be

0%
Bifocals with power -0.5 D and additional +3.5 D
0%
bifocals with power -1.0 D and additional +3 D
0%
Concave with power 1.0 D
0%
convex with power 0.5 D

What do ray PE denote?

0%
Angle of incidence
0%
Incident ray
0%
Refracted ray
0%
Reflected ray

Explanation

Ray

$PE$ is the incident ray which is getting refracted at the face

$AB$ of the prism.

Ray

$EF$ is the refracted ray which is getting refracted again at the face

$AC$ of the prism.

Ray

$FS$ is the emergent ray coming out the face

$AC$ of the prism.

So option B is the answer.

Ray FS is the refracted ray

0%
True
0%
False

Name the angle of prism.

0%
$\angle$ $A$
0%
$\angle$ $B$
0%
$\angle$ $C$
0%
$\angle$ $D$

Explanation

Angle of prism is the angle between the two surface of the prism from which the light enters into the prism and from the light goes out after refraction.

$\angle$

$A$ Represent angle of prism. Hence option A.
2004505_1954223_ans_c897215f729547679cc052c54ef03bc4.png

CBSE Questions for Class 10 Physics Human Eye And Colorful World Quiz 16 - 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 16 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

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