CBSE Class 10 Physics The Human Eye and the Colourful World Worksheet Set 07

This printable worksheet (Set 07) for CBSE Class 10 Physics Chapter 10, The Human Eye and the Colourful World, has eleven fill-in-the-blank questions and four numerical problems on the power of corrective lenses, in a short print-friendly format. It practises the parts of the eye, the near point and the range of vision, accommodation, dispersion and scattering, and the correction of myopia and hypermetropia, and every question has its answer below it.

Fill in the Blanks

Question 1. Fill in the blank: The coloured diaphragm between the cornea and the lens is the ______.
Answer:
iris
In simple words: The iris is the coloured ring of muscle.

Teacher's Note:
1. It controls the size of the pupil.
2. Do not confuse with the pupil.

Question 2. Fill in the blank: The middle point of the iris has a hole, which is called the ______.
Answer:
pupil
In simple words: The hole in the iris is the pupil.

Teacher's Note:
1. The size of the pupil is changed by the iris.
2. It controls the light entering the eye.

Question 3. Fill in the blank: The screen on which the image is formed by the lens system of the human eye is called the ______.
Answer:
retina
In simple words: The retina is the screen of the eye.

Teacher's Note:
1. The image on it is real and inverted.
2. It contains rods and cones.

Question 4. Fill in the blank: For a young adult with normal vision, the least distance of distinct vision is ______.
Answer:
25 cm
In simple words: About 25 cm for a normal eye.

Teacher's Note:
1. It is also the near point.
2. It increases with age.

Question 5. Fill in the blank: The closest distance at which the eye can focus clearly is called the ______ of the eye.
Answer:
near point
In simple words: The near point is the closest clear distance.

Teacher's Note:
1. Its value is about 25 cm for a normal eye.
2. For a hypermetropic eye it is farther away.

Question 6. Fill in the blank: For a normal eye, the range of vision is from ______.
Answer:
25 cm to infinity
In simple words: A normal eye sees clearly from 25 cm to infinity.

Teacher's Note:
1. The lower limit is the near point.
2. The upper limit is the far point.

Question 7. Fill in the blank: A person is short-sighted if his eyeball is too ______.
Answer:
long
In simple words: A long eyeball causes myopia.

Teacher's Note:
1. The image forms in front of the retina.
2. A concave lens corrects it.

Question 8. Fill in the blank: The ability of the eye to focus both near and distant objects, by adjusting its focal length, is called the ______ of the eye.
Answer:
power of accommodation
In simple words: The eye's power to focus near and far.

Teacher's Note:
1. It is done by the ciliary muscles.
2. It decreases with age.

Question 9. Fill in the blank: The splitting of white light into its component colours is called ______.
Answer:
dispersion
In simple words: Splitting of white light into colours is dispersion.

Teacher's Note:
1. A prism shows dispersion.
2. The coloured band is the spectrum.

Question 10. Fill in the blank: ______ of light causes the blue colour of the sky and the reddening of the Sun at sunrise and sunset.
Answer:
Scattering
In simple words: Scattering gives the blue sky and the red Sun.

Teacher's Note:
1. Short wavelengths are scattered most.
2. The Sun reddens as blue is scattered away.

Question 11. Fill in the blank: Sunlight comprises ______ colours.
Answer:
7 (seven)
In simple words: Sunlight has seven colours: VIBGYOR.

Teacher's Note:
1. Violet, indigo, blue, green, yellow, orange, red.
2. These form the spectrum.

Numerical Problems

Question 12. The 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 defect?
Answer:
Myopia is corrected by a concave lens.
The lens must form a virtual image at the far point for a very distant object, so u = −∞ and v = −80 cm. 1/f = 1/v − 1/u = −1/80, so f = −80 cm = −0.8 m.
P = 1/f = 1/(−0.8) = −1.25 D.
In simple words: A concave lens of −1.25 D is needed.

Teacher's Note:
1. Far point decides the focal length.
2. Convert to metres before finding power.

Question 13. The far point of a myopic person is 120 cm in front of the eye. What is the nature and power of the lens required to correct the problem?
Answer:
The person has myopia, so a concave lens is used.
f = −120 cm = −1.2 m.
P = 1/f = 1/(−1.2) = −0.83 D.
In simple words: A concave lens of about −0.83 D is required.

Teacher's Note:
1. The image is in front of the retina in myopia.
2. Round the answer to two decimals.

Question 14. The near point of a hypermetropic eye is 0.8 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.
Answer:
A convex lens is needed.
u = −25 cm = −0.25 m and v = −0.8 m. 1/f = 1/v − 1/u = −1/0.8 + 1/0.25 = −1.25 + 4 = 2.75 per m.
P = +2.75 D (f = 0.364 m = 36.4 cm).
In simple words: A convex lens of +2.75 D is required.

Teacher's Note:
1. Do the arithmetic in metres.
2. Use exact values and round only at the end.

Question 15. The near point of a hypermetropic eye is 1 m. What is the nature and power of the lens required to correct this defect? (Assume that the near point of the normal eye is 25 cm.)
Answer:
The defect is hypermetropia, so a convex lens is required.
u = −25 cm and v = −100 cm. 1/f = 1/v − 1/u = −1/100 + 1/25 = 3/100, so f = 33.3 cm.
P = 1/0.333 = +3.0 D.
In simple words: A +3 D convex lens is required.

Teacher's Note:
1. The virtual image must be formed at the defective near point.
2. Check that the sign of f is positive.