Human Eye and Colourful World Test

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Human Eye and Colourful World Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

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 $$40 \,cm$$.

A
$$2.5 D$$

B
$$-2.5 D$$

C
$$1.5 D$$

D
$$1.67 D$$

Solution
Question 2
1 / -0

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

A
$$(i)$$ and $$(ii)$$ only

B
$$(ii)$$ and $$(iii)$$ only

C
$$(i)$$ and $$(iii)$$ only

D
$$(i),(ii)$$ and $$(iii)$$

Solution
Question 3
1 / -0

Which phenomenon is responsible for the twinkling of stars?

A
Atmospheric reflection

B
Atmospheric refraction

C
Reflection

D
Total internal reflection

Solution

The atmospheric refraction is responsible for twinkling of stars. As the light from the stars enters the earth’s atmosphere it suffers refraction and cause the twinkling.
Question 4
1 / -0

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:

A
1.5 cm

B
5 cm

C
2.5 cm

D
1.67 cm

Solution

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 $$1.67\ cm$$
Question 5
1 / -0

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 :

A
$$\angle i + \angle e$$ = $$\dfrac {\angle P + \angle D}{2}$$

B
$$\angle i + \angle D = \angle P + \angle e$$

C
$$\angle i + \angle e = \angle P + \angle D$$

D
$$None\ of these$$

Solution

$$ \begin{array}{l} \text { we know, } \\ \qquad i+e=A+\delta \\ i \rightarrow \text { angle of incidence } \\ e \rightarrow \text { angle of emergence } \\ A \rightarrow \text { angle of prism } \\ \delta \rightarrow \text { angle of deviation } \\ \Rightarrow \quad \angle i+\angle e=\angle P+\angle D \end{array} $$
Question 6
1 / -0

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:

A
Short-sightedness

B
Long-sightedness

C
Cataract

D
Presbyopia

Solution

The optician has prescribed convex lenses for the child's spectacles because a convex lens is used to increase the converging power of the eye lens. This shows that child is suffering from the defect of vision called long-sightedness.

Long-sightedness is when the eye does not focus light on the retina (the light-sensitive layer at the back of the eye) properly. This may be because: the eyeball is too short. the cornea (transparent layer at the front of the eye) is too flat. the lens inside the eye is unable to focus properly.
Question 7
1 / -0

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.

A
$$50\ cm$$

B
$$\dfrac{200}{13}\ cm$$

C
$$\dfrac{200}{3}\ cm$$

D
$$25\ cm$$

Solution

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

$$ \text {Ans: } \dfrac{200}{3} \mathrm{~cm} $$
Question 8
1 / -0

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)

A
2D

B
3D

C
-2D

D
-3D
Question 9
1 / -0

The intensity of scattered light

A
$$I\propto { \lambda }^{ 4 }$$

B
$$I\propto { \lambda }^{ -4 }$$

C
$$I\propto { \lambda }^{ -2 }$$

D
$$I\propto { \lambda }^{ 2 }$$

Solution
Question 10
1 / -0

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?

A
23 cm

B
34.5 cm

C
18.4 cm

D
30 cm

Solution

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.

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Re-Attempt Weekly Quiz Competition

Human Eye and Colourful World Test - 61

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Human Eye and Colourful World Test - 61

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SHARING IS CARING

Question 1

$$2.5 D$$

$$-2.5 D$$

$$1.5 D$$

$$1.67 D$$

Solution

Question 2

$$(i)$$ and $$(ii)$$ only

$$(ii)$$ and $$(iii)$$ only

$$(i)$$ and $$(iii)$$ only

$$(i),(ii)$$ and $$(iii)$$

Solution

Question 3

Atmospheric reflection

Atmospheric refraction

Reflection

Total internal reflection

Solution

Question 4

1.5 cm

5 cm

2.5 cm

1.67 cm

Solution

Question 5

$$\angle i + \angle e$$ = $$\dfrac {\angle P + \angle D}{2}$$

$$\angle i + \angle D = \angle P + \angle e$$

$$\angle i + \angle e = \angle P + \angle D$$

$$None\ of these$$

Solution

Question 6

Short-sightedness

Long-sightedness

Cataract

Presbyopia

Solution

Question 7

$$50\ cm$$

$$\dfrac{200}{13}\ cm$$

$$\dfrac{200}{3}\ cm$$

$$25\ cm$$

Solution

Question 8

2D

3D

-2D

-3D

Question 9

$$I\propto { \lambda }^{ 4 }$$

$$I\propto { \lambda }^{ -4 }$$

$$I\propto { \lambda }^{ -2 }$$

$$I\propto { \lambda }^{ 2 }$$

Solution

Question 10

23 cm

34.5 cm

18.4 cm

30 cm

Solution

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