CBSE Class 10 Science Light, Human Eye and Electricity Important Questions Set 03

Answer: The change in the speed of light as it passes from one medium to another.

Teacher's Note:
1. Refraction occurs at the boundary between two media.
2. The speed changes because the media differ in optical density.

Answer: Its speed increases and the ray bends away from the normal at the interface.

Teacher's Note:
1. Air is optically rarer than glass.
2. A ray entering a rarer medium bends away from the normal.

Answer: It depends on the difference between the speeds of light in the two media; a larger difference gives greater bending.

Teacher's Note:
1. Read the fourth sentence of the case.
2. The bending is related to the ratio of the speeds, i.e. the refractive index.

Answer: The difference in speed is larger for air and glass (1 × 10⁸ m/s) than for air and water (0.75 × 10⁸ m/s).

Teacher's Note:
1. Subtract the speeds to compare the differences.
2. Glass has the higher refractive index of the two.

Answer: (c) real. Only real images, formed by actual meeting of rays, can be obtained on a screen.

Teacher's Note:
1. A virtual image cannot be caught on a screen.
2. Real images are formed by converging rays.

Answer: (a) Red. Red has the longest wavelength and the lowest refractive index in glass, so it bends the least.

Teacher's Note:
1. Violet bends the most.
2. The spectrum is ordered by deviation from red to violet.

Answer: (d) 25 W. P = V²/R, so halving the voltage reduces power to one-fourth, i.e. 100/4 = 25 W.

Teacher's Note:
1. Resistance stays the same, so power varies as the square of voltage.
2. Half the voltage gives one-quarter of the power.

Answer: (b) IR². Power is given by VI, I²R and V²/R; IR² is not a power expression.

Teacher's Note:
1. Combine P = VI with Ohm's law to get the other forms.
2. Check the dimensions of IR² if in doubt.

Answer: (d) retina. The eye lens focuses light on the retina, where light-sensitive cells send signals to the brain.

Teacher's Note:
1. The retina acts like the screen of a camera.
2. The cornea only does most of the bending of light.

Answer: (c) 25 cm. This near point of the normal eye is about 25 cm.

Teacher's Note:
1. The distance is quoted in centimetres.
2. It is also called the near point.

Answer: (c) ciliary muscles. Ciliary muscles change the curvature of the lens and so its focal length.

Teacher's Note:
1. This adjustment is called accommodation.
2. The iris controls the size of the pupil, not the focal length.

Answer: (b) Convex. A hypermetropic eye has too little converging power, so a convex lens is used.

Teacher's Note:
1. The eye cannot see near objects clearly.
2. A convex lens converges the rays before they enter the eye.

Answer: When sunlight strikes particles in the atmosphere, they absorb and re-emit the light in all directions; this is scattering. The colour of scattered light depends on the size of the scattering particles. Very fine particles scatter mainly blue light of short wavelength, larger particles scatter longer wavelengths, and very large particles scatter all colours and so appear white.

Teacher's Note:
1. Scattering depends on particle size and wavelength.
2. Shorter wavelengths scatter more strongly.

Answer:

1. The blue colour of the sky.

2. The reddening of the sun at sunrise and sunset.

(Also: the Tyndall effect, the white colour of clouds.)

Teacher's Note:
1. Both involve sunlight passing through the atmosphere.
2. The Tyndall effect is scattering by colloidal particles.

Answer: f = +30 cm, u = −24 cm.
\( \dfrac{1}{v} = \dfrac{1}{30} + \dfrac{1}{24} = \dfrac{9}{120} \), so v = +13.3 cm (behind the mirror).
m = −v/u = 13.3/24 = +0.56, so the image height is 4.5 × 0.56 = 2.5 cm.
The image is virtual, erect and diminished.

Teacher's Note:
1. Use the mirror formula with the sign convention.
2. A convex mirror always gives a virtual image behind it.

Answer: It is the ability of the eye lens to adjust its focal length by the action of the ciliary muscles so that the image of objects at different distances is formed on the retina. The eye lens becomes thicker for near objects and thinner for distant objects.

Teacher's Note:
1. Accommodation helps us focus on near and far objects.
2. The limit of this power fixes the near and far points.

Answer: The near point is the nearest distance at which the eye can see an object clearly, about 25 cm. The far point is the farthest point up to which the eye can see clearly, and for a normal eye it is at infinity.

Teacher's Note:
1. The range of clear vision is from 25 cm to infinity.
2. Defects of vision change these points.

Answer: The ciliary muscles can make the eye lens only so thick; for objects closer than 25 cm the focal length cannot be reduced enough. The image is then formed behind the retina and the eye feels strain.

Teacher's Note:
1. The limit is set by the muscles and the lens curvature.
2. The near point is about 25 cm for a normal eye.

Answer: Starlight undergoes atmospheric refraction as it passes through layers of air of changing optical density. This changes the apparent position of the star and the amount of light entering the eye slightly from moment to moment, and since a star is a point source, it appears to twinkle.

Teacher's Note:
1. The air layers are constantly moving.
2. A point source shows the fluctuation clearly.

Answer: Planets are much closer to us and can be taken as extended sources made of many point sources. The variations in light from the individual points cancel out on average, so the total light entering the eye stays almost steady.

Teacher's Note:
1. The difference lies in the size of the source.
2. A star is a point source, while a planet is extended.

Answer: At sunrise the sun's rays pass through a thicker layer of atmosphere to reach us. Most of the blue light and shorter wavelengths are scattered away, so the light that reaches the eye is mainly red.

Teacher's Note:
1. The path through the atmosphere is longest at sunrise and sunset.
2. Red light is scattered the least.

Answer: The image distance remains the same because the retina is at a fixed distance from the lens. The ciliary muscles change the focal length of the lens so that the image always forms on the retina.

Teacher's Note:
1. The retina cannot move.
2. The focal length adjusts instead.

Answer: Space has almost no atmosphere, so there are no particles to scatter sunlight. Without scattering, no light reaches the astronaut's eyes from the sky, and it looks dark.

Teacher's Note:
1. The blue sky is caused by scattering in the atmosphere.
2. The astronaut is above the atmosphere.

Answer: The lens should form a virtual image of a distant object at the far point of the eye, 80 cm away. For a distant object u = −∞, v = −80 cm, so f = −80 cm = −0.8 m.
P = 1/f = 1/(−0.8) = −1.25 D.
The lens is a concave lens of power −1.25 D.

Teacher's Note:
1. Myopia needs a diverging lens.
2. The focal length equals the far point distance with a negative sign.

Answer: Atmospheric refraction is the bending of light by the earth's atmosphere, whose optical density changes with height, so a ray follows a curved path. Examples: (1) twinkling of stars, caused by changing refraction of starlight; (2) the sun is visible about 2 minutes before actual sunrise and about 2 minutes after actual sunset, which lengthens the day; (3) the flattened look of the sun's disc near the horizon.

Teacher's Note:
1. The air near the ground is denser than the air above.
2. Rays bend towards the normal as they enter denser layers.

Answer: (a) For light of a given colour passing through a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant: \( \dfrac{\sin i}{\sin r} = n \).
(b) The refractive index of 2.42 means that the speed of light in diamond is 1/2.42 of its speed in vacuum, i.e. about 3 × 10⁸/2.42 ≈ 1.24 × 10⁸ m/s.

Teacher's Note:
1. The constant is the refractive index of the second medium relative to the first.
2. The refractive index compares the speed of light in vacuum with that in the medium.

Answer: f = −20 cm, v = −15 cm (virtual image on the same side).
\( \dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f} = -\dfrac{1}{15} + \dfrac{1}{20} = -\dfrac{1}{60} \)
So u = −60 cm, i.e. 60 cm in front of the lens.
m = v/u = (−15)/(−60) = 0.25, so h_i = 0.25 × 6 = 1.5 cm, virtual and erect.

Teacher's Note:
1. A concave lens always forms a virtual image on the same side as the object.
2. Use the lens formula 1/v − 1/u = 1/f.

Answer: Dispersion is the splitting of white light into its component colours (VIBGYOR) on passing through a prism. The cause is that different colours travel at different speeds in glass, so each has a different refractive index and bends by a different angle; violet bends the most and red the least.
Recombination: a second identical prism is placed upside down next to the first. The colours coming out of the first prism fall on the second, which bends them back so that they emerge together as white light.

Teacher's Note:
1. Dispersion is due to different bending of different wavelengths.
2. The second prism must be inverted for the colours to recombine.