CBSE Class 7 Science Advanced Physics Light Short Answer Questions Set 02

Short Answer Questions

Question 1. Why does not an aeroplane or a bird cast a shadow when they are at greater heights?
Answer:
The Sun is the source, the aeroplane is the opaque object and the Earth is the screen. When the aeroplane is low, the screen is near and a clear shadow forms. As it goes higher, the distance to the screen grows, the umbra becomes smaller and smaller and finally vanishes, so no shadow is seen on the ground.
In simple words: As the object goes far above, its shadow fades away.

Teacher's Note:
1. Shadow size depends on object-to-screen distance.
2. The Sun is a large (extended) source.

Question 2. Why cannot a mirror whose surface is rough produce a clear image?
Answer:
A rough surface reflects the light falling on it in many different directions (irregular or diffuse reflection), so the reflected rays do not form a definite image. A smooth surface gives regular reflection, where parallel rays stay parallel, and therefore a clear image.
In simple words: Rough surfaces scatter light; smooth ones reflect it in order.

Teacher's Note:
1. Compare regular and diffuse reflection.
2. Laws hold at every point even on rough surfaces.

Question 3. A clever student has used the concept of formation of shadow to determine the height of a long pole (OA). There was an electric bulb at the top of the pole. The student placed an object of length 1 m at a height of 2 m from the ground as shown in the figure. The size of the shadow formed on the ground is 2 m. Determine the length of the pole.

Answer:
Let OD = x. Triangles OAB and ODC are similar, so \( \dfrac{OA}{OD} = \dfrac{AB}{DC} \).
DC = ½ EC = 0.5 m and AB = ½ FB = 1 m.
\( \dfrac{x+2}{x} = \dfrac{1}{0.5} = 2 \), so x + 2 = 2x and x = 2 m.
Length of the pole OA = 2 + 2 = 4 m.
In simple words: Similar triangles show the bulb is 2 m above the object, so the whole pole is 4 m.

Teacher's Note:
1. Use similar triangles.
2. Check: shadow/object = 2 = OA/OD = 4/2.

Question 4. A pin hole camera of width 1 m is focussed to an object at a distance 30 m from it. If the height of the object is 3 m, what is the height of the image?
Answer:
Magnification \( m = \dfrac{h_i}{h_o} = \dfrac{v}{u} \), with v = 1 m (width of the camera) and u = 30 m.
\( \dfrac{h_i}{3} = \dfrac{1}{30} \)
h_i = 3/30 = 0.1 m (10 cm).
In simple words: The image is 30 times smaller than the object.

Teacher's Note:
1. The camera width is the image distance.
2. The image is inverted.

Question 5. Jim determined the number of images formed by two plane mirrors inclined at a certain angle from a ray diagram shown below. Can you determine the number of images formed by two plane mirrors in the arrangement shown below?

Answer:
At B the ray makes 40° with the normal, so it makes 50° with the mirror AB: ∠ABC = 50°.
At C the ray makes 20° with the normal to AC, so it makes 70° with the mirror AC: ∠ACB = 70°.
In triangle ABC, ∠CAB = θ = 180° − 50° − 70° = 60°.
360°/θ = 360°/60° = 6 (even), so n = 6 − 1 = 5 images.
In simple words: Find the angle between the mirrors from the triangle, then use the image formula.

Teacher's Note:
1. Angle with the mirror = 90° − angle with the normal.
2. The angle sum of a triangle is 180°.

Question 6. Rama watched in a discovery channel that in outer space, at greater heights above the surface of the Earth where atmosphere does not exist, the sky appears completely dark. Explain what would be the reason for this.
Answer:
The sky looks bright and blue on the Earth because the air, dust and water particles scatter sunlight in all directions into our eyes. At great heights there is no atmosphere, so there are no particles to scatter the light. No light reaches the eyes from the sides, and the sky appears dark.
In simple words: No air, no scattering, so the sky looks black.

Teacher's Note:
1. Scattering needs particles in the air.
2. Stars are still visible because they send light directly.

Question 7. Express 100 W in candlepower.
Answer:
1 W = 683 lumen and 1 lumen = 12.56 candle power.
100 W = 100 × 683 lumen = 68,300 lumen.
68,300 × 12.56 = 8,57,848 candle power (approximately).
So the source is about 8,57,848 times as intense as a standard candle.
In simple words: Convert watt to lumen, then lumen to candle power.

Teacher's Note:
1. Multiply 100 × 683 × 12.56.
2. Keep the conversion numbers handy.

Question 8. A physics teacher asked a student to find how much time the light takes to reach the Earth if the distance between the Sun and the Earth is 150 × 10⁶ km. The teacher also gives an extra input that the velocity of light is 3 × 10⁸ m s⁻¹. What would be the answer of the student?
Answer:
Distance d = 150 × 10⁶ km = 150 × 10⁹ m.
Time t = d/c = \( \dfrac{150 \times 10^{9}}{3 \times 10^{8}} \) = 500 s.
500 s ÷ 60 = 8.33 minutes (about 8 minutes 20 seconds).
In simple words: Sunlight takes about 8 minutes 20 seconds to reach us.

Teacher's Note:
1. Change km to m first.
2. Time = distance ÷ speed.

Question 9. Why is a convex mirror used as a rear-view mirror in vehicles? Explain.
Answer:
A convex mirror always forms an erect, virtual and diminished image, and it has a very wide field of view. So it shows objects spread over a large area behind the vehicle in a small mirror, and the driver can see the traffic coming from behind.
In simple words: A convex mirror shows a big area in a small space.

Teacher's Note:
1. Mention field of view and diminished image.
2. Plane mirrors would show a narrower area.

Question 10. A physics student proved that 'sunlight consists of different colours' and even gives two examples to support this. Explain.
Answer:
1. Rainbow: sunlight passing through raindrops is split into seven colours.
2. Prism: when sunlight passes through a glass prism, a band of seven colours is seen on a screen.
Other examples: soap bubbles and a compact disc show many colours in sunlight.
Hence sunlight (white light) is a mixture of colours.
In simple words: Rainbows and prisms split sunlight into colours, so sunlight is a mixture of them.

Teacher's Note:
1. Name the phenomenon: dispersion.
2. Give at least two examples.

Question 11. Explain the formation of real and virtual images. Discuss what kind of images are formed in plane, convex and concave mirrors.
Answer:
A real image is formed when reflected rays actually meet (a convergent beam). It can be caught on a screen and is inverted.
A virtual image is formed when reflected rays only appear to meet when extended backwards (a divergent beam). It cannot be caught on a screen and is erect.
Plane mirror: virtual, erect, same size, laterally inverted.
Convex mirror: virtual, erect and diminished for every object position.
Concave mirror: real, inverted images (smaller, equal or larger) when the object is beyond F; virtual, erect, magnified image when the object is between P and F.
In simple words: Real images are made by meeting rays and virtual images by apparent meeting.

Teacher's Note:
1. Learn this together with the table of concave mirror images.
2. Only a concave mirror makes real images.

Question 12. In the head lights of a car, what kind of mirrors are used? What is the use of mirror in the car head lights? Explain.
Answer:
Concave mirrors are used. The bulb is placed at the principal focus of the mirror. Rays from the focus become parallel after reflection, so the light goes out as a strong parallel beam that travels a long way and lights the road ahead.
In simple words: The bulb sits at the focus, so a concave mirror sends its light out as a parallel beam.

Teacher's Note:
1. Rays through F become parallel.
2. The same idea is used in torches and searchlights.

Question 13. (a) Vishal wrote the word 'LIGHT' on a paper and placed it in front of a plane mirror (as shown in the figure) and noticed how the letters of the word appear in the mirror. Write down what he noticed in the mirror. (b) What is his observation about the nature of the image? And through what angle did the image get inverted when compared to the object?

Answer:
(a) Because the mirror is below the paper, the image looks upside-down: each letter appears with its top and bottom interchanged, and the letters stay in the same left-to-right order.

(b) The image is virtual (it cannot be caught on a screen) and laterally inverted. As in the book's solution, the image is inverted through 180° compared with the object.
In simple words: The mirror flips the word upside-down and the image is virtual.

Teacher's Note:
1. Lateral inversion swaps the side facing the mirror.
2. A virtual image cannot be caught on a screen.

Question 14. A bus driver is observing the image of a car in his rear view (side) mirror. How does the size of the image change when the car approaches the bus?
Answer:
The rear-view mirror is convex. When the car is far away, its image is very small and formed near the focus. As the car comes nearer, the image moves away from the focus towards the object side and its size increases gradually, though it always remains smaller than the car.
In simple words: The image grows as the car gets closer.

Teacher's Note:
1. Convex mirror image is always diminished.
2. Size of image increases as u decreases.

Question 15. When the physics teacher has given a task to determine the focal length of a concave mirror, a clever boy in the class focussed the mirror to a distant object. He moved the screen in front of the mirror and caught a real image on a screen at a distance of 20 cm from the mirror and he determined the focal length. Where should the object be placed so that the object and the image are formed at the same position?
Answer:
Rays from a distant object are parallel and meet at the focus, so the image on the screen is at F: f = 20 cm.
The object and its image coincide when the object is at the centre of curvature C.
R = 2f = 2 × 20 = 40 cm, so the object should be placed 40 cm in front of the mirror.
In simple words: Put the object at twice the focal length.

Teacher's Note:
1. A distant object gives the focal length.
2. At C the image is as big as the object.