Question 1

Question 1The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due topresbyopiaaccommodationnear-sightednessfar-sightedness

Accepted Answer

accommodationReason — The ability of the eye lens to adjust its focal length is called accommodation.

Question 2

Question 2The human eye forms the image of an object at itscorneairispupilretina

Accepted Answer

retinaReason — The eye lens forms an inverted real image of the object on the retina.

Question 3

Question 3The least distance of distinct vision for a young adult with normal vision is about25 m2.5 cm25 cm2.5 m

Accepted Answer

25 cmReason — For human eye with normal vision, the near point or least distance of distinct vision is about 25 cm.

Question 4

Question 4The change in focal length of an eye lens is caused by the action of thepupilretinaciliary musclesiris

Accepted Answer

ciliary musclesReason— The ciliary muscles contract to make the lens thinner and relax to make it thicker, thus changing its curvature and hence focal length.

Question 5

Question 5A person needs a lens of power –5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?

Accepted Answer

Power is reciprocal of focal length.(i) Focal length of the lens required for correcting distant vision = −15.5-\dfrac{1}{5.5}−5.51​ = -0.181 m = -18.1 cm(ii) Focal length of the lens required for correcting near vision = 11.5\dfrac{1}{1.5}1.51​ = 0.667m = 66.7cm

Question 6

Question 6The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?

Accepted Answer

Concave lens is used to correct the problem of myopia.The image must be formed at far point.Given,Far point = 80 cm = 0.8 m∴ Power of lens required = −10.8-\dfrac{1}{0.8}−0.81​ = -1.25 dioptres.

Question 7

Question 7Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

Accepted Answer

Hypermetropia can be corrected by using a convex lens as shown in the diagram below:An object at 25 cm forms an image at the near point of hypermetropic eye.Given,Near point of hypermetropic eye = 1 m =100 cmObject distance, u = -25 cmImage distance, v = -100 cmAccording to the formula,⇒1v−1u=1f⇒1−100−1−25=1f⇒−1+4100=1f⇒3100=1f⇒f=1003 cm⇒f=13 mPower=1f=113∴Power=3 D\phantom{\Rightarrow} \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \[1 em] \Rightarrow \dfrac{1}{-100} - \dfrac{1}{-25} = \dfrac{1}{f} \[1 em] \Rightarrow \dfrac{-1 + 4}{100} = \dfrac{1}{f} \[1 em] \Rightarrow \dfrac{3}{100} = \dfrac{1}{f} \[1 em] \Rightarrow f = \dfrac{100}{3} \text{ cm} \[1 em] \Rightarrow f = \dfrac{1}{3} \text{ m} \[1 em] \text{Power} = \dfrac{1}{f} = \dfrac{1}{\dfrac{1}{3}} \[1 em] \therefore \text{Power} = 3 \text{ D}⇒v1​−u1​=f1​⇒−1001​−−251​=f1​⇒100−1+4​=f1​⇒1003​=f1​⇒f=3100​ cm⇒f=31​ mPower=f1​=31​1​∴Power=3 DHence, the power of the lens required to correct this defect is 3 dioptre.

Question 8

Question 8Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

Accepted Answer

When focusing on a nearby object, the ciliary muscles strain to adjust the eye lens, reducing its focal length. However, there's a limit to this adjustment. Beyond approximately 25 cm, the eye struggles to bring the image into clear focus, resulting in blurred vision for objects placed closer. This limitation is due to the eye's finite ability to accommodate and maintain sharp focus.

Question 9

Question 9What happens to the image distance in the eye when we increase the distance of an object from the eye?

Accepted Answer

The image distance in the eye remains the same because the image is always formed on retina. When we increase the distance of an object from the eye, only the lens changes its thickness to form the image of the object on retina.

Question 10

Question 10Why do stars twinkle?

Accepted Answer

Stars twinkle due to atmospheric refraction of starlight. As starlight passes through Earth's atmosphere, it undergoes continuous refraction in a medium of gradually changing refractive index, bending towards the normal. This bending causes the apparent position of the star to fluctuate slightly, leading to the twinkling effect. The dynamic and non-uniform conditions of the Earth's atmosphere, especially near the horizon, contribute to this phenomenon.

Question 11

Question 11Explain why the planets do not twinkle.

Accepted Answer

Planets do not twinkle because, unlike stars, they are seen as extended sources of light. When observing a planet, which appears as a disk rather than a point of light, the combined effect of a large number of point-sized sources averages out the total variation in the amount of light entering the observer's eye. This averaging effect nullifies the twinkling phenomenon, making planets appear steady in brightness.

Question 12

Question 12Why does the sky appear dark instead of blue to an astronaut?

Accepted Answer

The molecules of air and other fine particles in the atmosphere have size smaller than the wavelength of visible light. These are more effective in scattering light of shorter wavelengths i.e blue. When sunlight passes through the atmosphere, the fine particles in air scatter the blue colour (shorter wavelengths) more strongly than red. The scattered blue light enters our eyes and the sky appears blue. But there are no particles, hence no scattering of light in space. Therefore, the sky appear dark to an astronaut.

Question 13

Question 1What is meant by power of accommodation of the eye?

Accepted Answer

Power of accommodation of the eye is the ability of the eye lens to focus near and far object clearly on the retina by adjusting the thickness of lens (and hence focal length). The ciliary muscles attached to the lens help the thickening or thinning of lens hence changing its curvature.

Question 14

Question 2A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?

Accepted Answer

A myopic eye can be corrected by using a concave lens of suitable power. This will restore proper vision.

Question 15

Question 3What is the far point and near point of the human eye with normal vision?

Accepted Answer

The minimum distance, at which objects can be seen most distinctly without strain, is called the least distance of distinct vision or the near point of the eye. For human eye with normal vision, the near point is about 25 cm. The farthest point upto which the eye can see objects clearly is called the far point of the eye. It is infinity for a normal eye.

Question 16

Question 4A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?

Accepted Answer

The student has difficulty reading the blackboard while sitting in the last row, this means he cannot see distant objects clearly. Therefore, child could be suffering from myopia. Myopia can be corrected by using a concave lens of suitable power.

The Human Eye and the Colourful World

Exercises

Intext Questions 1