Human Eye and Colourful World Test

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Human Eye and Colourful World Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

A short-sighted person can see objects most distinctly at a distance of $$16 cm$$. If he wears spectacles at a distance of $$1 cm$$ from the eye, then their focal length to see at a distance of $$26 cm$$ distinctly is

A
$$25cm, convex$$

B
$$25 cm, concave$$

C
$$37.5 cm, convex$$

D
$$37.5 cm, concave$$
Solution
The farthest that the person can see is 16 cm. It means that rays coming from an object, which is 16 cm away from the eye, can be focused properly. When the person wears spectacles, the eyes don't see the objects directly, instead they see only the image of the objects formed by the lenses of the spectacles. So, in order to see some thing 26 cm away we should make sure that the image of object formed by spectacles should be 16 cm away so that eye could focus the rays coming from the image.
Object is 26 cm away from eye or 25 cm away from the lens of the spectacles ($$ u = -25 cm$$)
Image is 16 cm from the eye or 15 cm away from the lens of the spectacles ($$ v = -15 cm$$)
(Taking the left side to the lens of spectacle to be positive)
Lens formula: $$ \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} $$
Substituting $$u$$ and $$v$$ in the lens equation:
$$ \dfrac{1}{-15} - \dfrac{1}{-25} = \dfrac{1}{f} $$
$$ \dfrac{-5}{75} + \dfrac{3}{75} = \dfrac{1}{f} $$
$$ \dfrac{-2}{75} = \dfrac{1}{f} $$
$$\Rightarrow f = -75/2 = -37.5 cm $$
Focal length is negative & rays are coming from right to left. That means lens is concave.
Question 2
1 / -0

The near point and the far point of a child are at $$10\ cm$$ and $$100\ cm$$ respectively. If the retina is $$2\ cm$$ behind the eye lens, the range of the power of the eye lens is:

A
$$40\ D$$ to $$20\ D$$

B
$$60\ D$$ to $$51\ D$$

C
$$35\ D$$ to $$22\ D$$

D
$$55\ D$$ to $$25\ D$$

Solution

We know that power of a lens, $$p = \dfrac{100}{f} $$ D, where $$f$$ is focal length expressed in $$cm$$.
We will apply the lens formula to calculate the focal length ($$f$$).
Lens formula: $$ \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} $$
In the first case (when the object is placed at the near point and image is formed on the ratina),
$$v = 2 \space cm$$ and $$u = -10 \space cm$$
$$ \dfrac{1}{2} - \dfrac{1}{-10} = \dfrac{1}{f} $$
$$ \Rightarrow \dfrac{5}{10} + \dfrac{1}{10} = \dfrac{1}{f} $$
$$ \Rightarrow \dfrac{6}{10} = \dfrac{1}{f} $$
$$\Rightarrow f = \dfrac{10}{6} = \dfrac{5}{3} $$
Thus, power $$P = \dfrac{100}{f} = \dfrac{100\times 3}{5} = +60 \space D$$
In the second case (when the object is placed at the far point and image is formed on the ratina),
$$v = 2 \space cm$$ and $$u = -100 \space cm$$
$$ \dfrac{1}{2} - \dfrac{1}{-100} = \dfrac{1}{f} $$
$$ \Rightarrow \dfrac{50}{100} + \dfrac{1}{100} = \dfrac{1}{f} $$
$$ \Rightarrow \dfrac{51}{100} = \dfrac{1}{f} $$
$$\Rightarrow f = \dfrac{100}{51} $$
Thus, power $$P = \dfrac{100}{f} = \dfrac{100\times 51}{100} = +51 \space D$$
Therefore, the range of the power of the eye lens is $$60 \space D$$ to $$51 \space D$$
Question 3
1 / -0

A person can see clearly objects lying between $$25\ cm$$ and $$2\ m$$ from his eye. His vision can be corrected by using spectacles of power:

A
$$+0.25\ D$$

B
$$+0.5\ D$$

C
$$-0.25\ D$$

D
$$-0.5\ D$$

Solution

Here, $$u= 2m$$ ; object distance
$$v= \infty $$ ; image distance
Apply formula:
$$\dfrac{1}{v}-\dfrac{1}{u}= \dfrac{1}{f}$$
$$f$$ : focal length
$$\Rightarrow \dfrac{1}{\infty}-\dfrac{1}{2}= \dfrac{1}{f}$$
$$\Rightarrow f= -2m$$
Now, relation between power$$(P)$$ and focal length $$(f)$$-
$$P= \dfrac{1}{f}$$
$$\Rightarrow P= \dfrac{1}{-2}D$$
$$\Rightarrow P= -0.5\ D$$
Question 4
1 / -0

Match the elements of table I and II.
Table I Table II
a) Myopia i) Bifocal len
b) Hypermetropia ii) Cylindrical lens
c) Presbyopia iii) Concave lens
d) Astigmation iv) Convex lens

A
a - iii, b - iv, c- i, d -ii

B
a - iv, b - iii, c -i, d - ii

C
a - i, b - ii, c - iii, d - iv

D
a - ii, b - iv, c -i, d - iii

Solution

Myopia is inability to see distant objects as parallel rays from distant objects converge to a point in front of retina due to increased power of eye lens.Hence it is corrected by a concave lens of negative power to decrease net power.

Hyper metropia is inability to see near objects as diverging rays from near objects can not be converged to retina due to decreased power of eye lens.Hence a convex lens of positive power is used to increase power.

Presbyopia is caused by changes in the lens inside the eye. As people age, the lens becomes harder and less elastic, making it more difficult for the eye to focus on close objects.It is corrected by bifocal lenses.

Astigmatism is a type of refractive error in which the eye does not focus light evenly on the retina. It can be corrected by cylindrical lenses.
Question 5
1 / -0

A person wears glasses of power $$-2.0\ D$$. The defect of the eye and the far point of the person without the glasses will be

A
Nearsighted, 50 cm

B
Farsighted, 50cm

C
Nearsighted, 250 cm

D
Astigmatism, 50 m

Solution

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

$$=\dfrac{100}{-2}cm$$

$$= -50\ cm$$

The defect of eye will be nearsighted
Question 6
1 / -0

The near point of a person is 50cm and the far point is 1.5m. The spectacles required for reading purpose and for seeing distant objects are respectively:

A
$$+2D$$, $$-\dfrac{2}{3}$$D

B
+$$\dfrac{2}{3}$$D, $$-2D$$

C
$$-2D$$, +$$\dfrac{2}{3}$$D

D
$$-\dfrac{2}{3}$$D, $$2D$$

Solution

For reading purpose,

$$u = 50 cm\ ,v= 25 cm$$

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

$$\dfrac{1}{25}-\dfrac{1}{50}= \dfrac{1}{f}$$

$$f= 50cm$$

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

$$f$$ should be in metre

$$P = \dfrac{100}{50}$$

$$P= +2 D$$

For seeing distance objects,

$$u = 1.5 m, \ \ \ v = \infty $$

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

$$f= -\dfrac{3}{2}m$$

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

$$P= \dfrac{-2}{3}D$$
Question 7
1 / -0

A defective eye cannot see close objects clearly because their image is formed:

A
On the eye lens

B
Between eye lens and retina

C
On the retina

D
Beyond retina

Solution

A defective eye cannot see close objects clearly because its image is formed beyond the retina. This defect is known as farsightedness. Here rays from near objects could not be converted at the retina due to focal length of the eye lens could not be decreased, so, image is formed beyond the retina.
Question 8
1 / -0

When white light enters a prism, its gets splits into its constituent colours. This is because of:

A
high density of prism material

B
$$ \mu $$, which is different for different wavelengths

C
diffraction of light

D
interference of light

Solution

Refractive index of a medium is dependent on the wavelength of light. White light constitutes of many colours. According to snell's law,
$$\mu = \dfrac{\sin i}{\sin r}$$
As $$\mu$$ is different for different wavelengths, $$r$$ is different too, and hence each wavelength is deviated by a different amount. This leads to the splitting of colours in white light.
Question 9
1 / -0

At noon the sun appears white because:

A
light is least scattered.

B
all the colours of the white light are scattered away.

C
blue colour is scattered the most.

D
red colour is scattered the most.

Solution

At noon because the sun is overhead, the light is scattered the least and hence appears white. When it is overhead, it has lesser air to travel through and the scattering from dust and other particles is reduced if the distance to be travelled in air is reduced.
Question 10
1 / -0

A beam of light composed of red and green rays is incident obliquely at a point on the face of a rectangular glass slab. When coming out on the opposite parallel face, the red and green rays emerge from :-

A
Two points propagating in two different parallel directions

B
One point propagating in two different directions through slab

C
One point propagating in the same direction through slab

D
Two points propagating in two different non parallel directions

Solution

Red and green rays emerge from two points, propagating in two different parallel directions.

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Re-Attempt Weekly Quiz Competition

Table I	Table II
a) Myopia	i) Bifocal len
b) Hypermetropia	ii) Cylindrical lens
c) Presbyopia	iii) Concave lens
d) Astigmation	iv) Convex lens

Human Eye and Colourful World Test - 27

Result

Human Eye and Colourful World Test - 27

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SHARING IS CARING

Question 1

$$25cm, convex$$

$$25 cm, concave$$

$$37.5 cm, convex$$

$$37.5 cm, concave$$

Solution

Question 2

$$40\ D$$ to $$20\ D$$

$$60\ D$$ to $$51\ D$$

$$35\ D$$ to $$22\ D$$

$$55\ D$$ to $$25\ D$$

Solution

Question 3

$$+0.25\ D$$

$$+0.5\ D$$

$$-0.25\ D$$

$$-0.5\ D$$

Solution

Question 4

a - iii, b - iv, c- i, d -ii

a - iv, b - iii, c -i, d - ii

a - i, b - ii, c - iii, d - iv

a - ii, b - iv, c -i, d - iii

Solution

Question 5

Nearsighted, 50 cm

Farsighted, 50cm

Nearsighted, 250 cm

Astigmatism, 50 m

Solution

Question 6

$$+2D$$, $$-\dfrac{2}{3}$$D

+$$\dfrac{2}{3}$$D, $$-2D$$

$$-2D$$, +$$\dfrac{2}{3}$$D

$$-\dfrac{2}{3}$$D, $$2D$$

Solution

Question 7

On the eye lens

Between eye lens and retina

On the retina

Beyond retina

Solution

Question 8

high density of prism material

$$ \mu $$, which is different for different wavelengths

diffraction of light

interference of light

Solution

Question 9

light is least scattered.

all the colours of the white light are scattered away.

blue colour is scattered the most.

red colour is scattered the most.

Solution

Question 10

Two points propagating in two different parallel directions

One point propagating in two different directions through slab

One point propagating in the same direction through slab

Two points propagating in two different non parallel directions

Solution

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