Human Eye and Colourful World Test

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Human Eye and Colourful World Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

Why are red coloured lights are used in traffic signals to stop the vehicles?

A
minimum wavelength and less scatter.

B
maximum wavelength and more scatter.

C
maximum wavelength and less scatter.

D
less wavelength and less scatter.

Solution

We know among the seven colours of white light, red colour has maximum wavelength. That is why red colour is least scattered by atmospheric dust and other particles. Hence red colour is used to indicated any danger sign.
Question 2
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A professor reads a greeting card on his $$50th$$ birthday with $$+2.5D$$ glasses keeping the card $$25\ cm$$ away. $$10$$ years later he reads the greeting card with the same glasses keeping the card $$50\ cm$$ away. What power should he wear now?

A
$$2\ D$$

B
$$0.5\ D$$

C
$$2.25\ D$$

D
$$4.5\ D$$
Question 3
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The image formed by the eye lens on the retina is :

A
Real, upright and enlarged

B
Real, upright and diminished

C
Real, inverted and diminished

D
Virtual, inverted and diminished

Solution

The eye lens forms an inverted, real, and diminished image of the object on the retina.
Thus option $$C$$ is correct.
Question 4
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Identify the wrong description of the given figures.

A
1 represents far sightedness

B
2 is correction for short sightedness

C
3 represent far sightedness

D
4 represents correction for far sightedness

Solution

In short-sightedness, Images of objects form before the retina, and for its correction diverging lenses are used.
In far-sightedness, Images of objects form behind the retina and for its correction converging lenses are used.
Here fig 1; showing short-sighted eyes. Hence option A is wrong.
Question 5
1 / -0

In the absence of atmosphere, the sky appears :

A
coloured

B
blue

C
indigo

D
black

Solution

It would appear black as there is no atmosphere no rays are refracted and hence light would travel in straight line .
It is because of atmosphere the light refracts and reflects and the blue light reaches our eye and hence the sky appears to be in blue.
Question 6
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The muscles of a normal eye are least strained when the eye is focused on an object:

A
Far away from the eye

B
Very close to the eye

C
At about $$25\ cm$$ from the eye

D
At about $$1\ m$$ from the eye

Solution

The eye lens is composed of fibrous, jelly-like material. Its curvature can be modified to some extent by the ciliary muscles. The change in the curvature of the eye lens can thus change its focal length. When the muscles are relaxed, the lens becomes thin. Thus, its focal length increases. This enables us to see distant objects clearly. When we are looking at objects closer to the eye, the ciliary muscles contract. This increases the curvature of the eye lens. The eye lens then becomes thicker. Consequently, the focal length of the eye lens decreases. This enables us to see nearby objects clearly.
So, the muscles of a normal eye are least strained when the eye is focused on an object far away from the eye as the muscles are relaxed.
Question 7
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The ciliary muscles of eye control the curvature of the lens in the eye and hence and alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained, the curvature of lens increases. That means, radius of curvature decreases and focal length decreases. For a clear vision, the image must be on the retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye lens. It is about $$25\ cm$$ for a grown up person.
A person can theoretically have clear vision of an object situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown up person, minimum distance of the object should be around $$25\ cm$$.
A person suffering from eye defects uses spectacles (eye glass). The function of lens of spectacles is to form the image of the objects within the range in which the person can see clearly. The image of the spectacle lens becomes object for the eye lens and whose image is formed on the retina.
The number of spectacle lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle lens is equal to the numerical value of the power of lens with sign. For example, if power of the lens required is $$+3D$$ (converging lens of focal length $$100/3\ cm$$), then number of lens will be $$+3$$.
For all the calculations required, you can used the lens formula and lensmaker's formula. Assume that the eye lens is equiconvex lens. Neglect the distance between the eye lens and the spectacle lens.
A near sighted man can clearly see objects only upto a distance of $$100 cm$$ and not beyond this. The number of the spectacle lenses necessary for the remedy of this defect will be

A
$$+1$$

B
$$-1$$

C
$$+3$$

D
$$-3$$

Solution

The power of the lens should be such that the image of an object at infinity is obtained at a distance of $$100 \ cm = 1 \ m$$ by the lens.

Hence, using the lens' formula

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

Let, $$u= -\infty$$ (object at infinity)
$$v=-1 \ m$$

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

Thus Power $$P=\dfrac{1}{f}=-1 \ D$$

Hence, the number of the spectacle lens will be $$-1$$.
Question 8
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Directions For Questions

Fill in the blank spaces by choosing the correct words from the list given below:
List : two, base, dispersion, prism, deviation

...view full instructions

When the refraction takes place through a glass prism the emergent ray always bends towards the .......... of prism.

A
base

B
two

C
dispersion

D
deviation

Solution

Base
On the first surface, the incident ray bends towards the normal i.e. towards the base and splits into different color bands. Upon striking the second surface each band is deviated away from the normal which is towards the base and hence the emergent ray is bent towards the base of the prism.
Question 9
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A normal eye is not able to see objects closer than 25 cm because:

A
the focal length of the eye is 25 cm

B
the distance of the retina from the eye-lens is 25 cm

C
the eye is not able to decrease the distance between the eye-lens and retina beyond a limit

D
the eye is not able to decrease the focal length beyond a limit

Solution

To see an object comfortably and distinctly, you must hold it at a distance 25 cm or more than 25 cm from the eyes. The minimum distance at which objects can be seen most distinctly without strain is called the least distance of distinct vision. It is also called the near point of the eye. For a young adult with normal vision, the near point is about 25 cm. When an object kept at a distance less than 25 cm then it will look blur because focal length of eyes will not decrease more than 25 cm. the ciliary muscules unable to contract after a certain limit
Thus, option $$D$$ is correct.
Question 10
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Hypermetropia is

A
same thing as shortsightedness

B
same thing as long sightedness

C
sane as cataract

D
another name of night blindness

Solution

Hyperopia or Far-sightedness is the condition where person can see the far objects clearly but cannot focus on near objects. This occurs since the image of nearby objects is formed behind retina.

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Human Eye and Colourful World Test - 32

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Human Eye and Colourful World Test - 32

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SHARING IS CARING

Question 1

minimum wavelength and less scatter.

maximum wavelength and more scatter.

maximum wavelength and less scatter.

less wavelength and less scatter.

Solution

Question 2

$$2\ D$$

$$0.5\ D$$

$$2.25\ D$$

$$4.5\ D$$

Question 3

Real, upright and enlarged

Real, upright and diminished

Real, inverted and diminished

Virtual, inverted and diminished

Solution

Question 4

1 represents far sightedness

2 is correction for short sightedness

3 represent far sightedness

4 represents correction for far sightedness

Solution

Question 5

coloured

blue

indigo

black

Solution

Question 6

Far away from the eye

Very close to the eye

At about $$25\ cm$$ from the eye

At about $$1\ m$$ from the eye

Solution

Question 7

$$+1$$

$$-1$$

$$+3$$

$$-3$$

Solution

Question 8

base

two

dispersion

deviation

Solution

Question 9

the focal length of the eye is 25 cm

the distance of the retina from the eye-lens is 25 cm

the eye is not able to decrease the distance between the eye-lens and retina beyond a limit

the eye is not able to decrease the focal length beyond a limit

Solution

Question 10

same thing as shortsightedness

same thing as long sightedness

sane as cataract

another name of night blindness

Solution

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