MCQ Questions for 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

0%
$\dfrac{4}{3}$
0%
$\sqrt{2}$
0%
$1.5$
0%
$\sqrt{3}$

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

0%
$40\ D$
0%
$4\ D$
0%
$-2.5\ D$
0%
$0.25\ D$

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

0%
Myopia
0%
Hypermetropia
0%
cataract
0%
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

$12\ \textit{metres}$

0%
$-3/4\ D$
0%
$3\ D$
0%
$-1/4\ D$
0%
$-4\ D$

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

$25\, cm$

0%
$+ 3.0\, D$
0%
$+ 0.125\, D$
0%
$- 3.0\, D$
0%
$+ 4.0\, D$

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.

0%
True
0%
False

In a glass prism, spectrum is produced due to :

0%
refraction
0%
dispersion
0%
scattering
0%
diffraction

The speeds of red light and yellow light are exactly same

0%
in vacuum but not in air
0%
in air but not in vacuum
0%
in vacuum as well as in air
0%
neither in vacuum nor in air.

A normal eye is not able to see objects closer than

$25 \ cm$ because?

0%
The focal length of the eye is $25 \ cm$
0%
The distance of the retina from the eye-lens is $25 \ cm$
0%
The eye is not able to decrease the distance between the eye-lens and the retina beyond a limit
0%
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

2114678_1717448_ans_02e860be1214482a8796ddbe6ea5e233.png

One can not see through fog because:

0%
fog absorbed light
0%
light is scattered by the droplets in fog
0%
light surfers total reflection by the droplets in the fog
0%
the refractive index of fog is in infinity

The rising and setting of sun appear red because of :

0%
refraction
0%
reflection
0%
defraction
0%
scattering

For a person suffering from myopia, the image of the object is formed in front of the retina.

0%
True
0%
False

Abnormalities in the normal vision of the eye are called defects of vision.

0%
True
0%
False

Short-sightedness is caused due to elongation of the eyeball.

0%
True
0%
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.

0%
True
0%
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.

0%
True
0%
False

A rainbow can be seen in the sky

0%
When the sun is in front of you.
0%
When the sun is behind you.
0%
When the sun is overhead.
0%
Only at the time of sun rise.

The eye defect represented by the figure is

0%
Myopia
0%
Hypermetropia
0%
Cataract
0%
Presbyopia

The color of the sky is blue during the daytime, red during sunset, and black at night. This is due to

0%
Scattering of light
0%
Small particles present in the atmosphere
0%
Atmospheric refraction
0%
All of the above.

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

0%
Myopia
0%
Hypermetropia
0%
Presbyopia
0%
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

0%
Planar lens
0%
Concave lens
0%
Convex lens
0%
Bifocal lens

The colored light that refracts most while passing through a prism is

0%
Yellow
0%
Violet
0%
Blue
0%
Red

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

0%
$+0.5\ D$
0%
$+1.0\ D$
0%
$+2.0\ D$
0%
$-1.0\ D$

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

0%
$\dfrac{10}{3}\ cm$
0%
$30\ cm$
0%
$15\ cm$
0%
$\dfrac{100}{3}\ cm$

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:

0%
$45\ cm$
0%
$-20\ cm$
0%
$-12\ cm$
0%
$30\ cm$

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

0%
$40\ D$
0%
$-4\ D$
0%
$-2.5\ D$
0%
$2.5\ D$

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

0%
Concave lens of $20\ cm$ focal length
0%
Convex lens of $20\ cm$ focal length
0%
Concave lens of $12\ cm$ focal length
0%
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

$20 \ cm$

2100534_1824164_ans_64dfd9a0a8884a74a63716e9f2e32e75.png

The hyper-metropia is a

0%
Short-side defect
0%
Long-side defect
0%
Bad vision due to old age
0%
None of these

When white light passes through the achromatic combination of prisms, then what is observed

0%
Only deviation
0%
Only dispersion
0%
Deviation and dispersion
0%
None of the above

Stars are twinkling due to

0%
Diffraction
0%
Reflection
0%
Refraction
0%
Scattering

Explanation

Stars twinkle due to variation in

$R.l$ of atmosphere.

Stars are twinkling due to refraction,

CBSE Questions for Class 10 Physics Human Eye And Colorful World Quiz 14 - MCQExams.com

0:0:1

Practice Class 10 Physics Quiz Questions and Answers

CBSE Questions for Class 10 Physics Human Eye And Colorful World Quiz 14 - MCQExams.com

0:0:1

Practice Class 10 Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.