CBSE Class 7 Science Advanced Physics Light Long Answer Questions Set 01

Long Answer Questions

Question 1. Write two phenomena which can be understood by rectilinear propagation of light.
Answer:
1. Formation of shadows: because light travels in straight lines, an opaque object stops the light and a dark region (shadow) forms behind it.
2. Eclipses: the Moon's shadow falling on the Earth gives a solar eclipse and the Earth's shadow falling on the Moon gives a lunar eclipse.
(Working of a pinhole camera is another example.)
In simple words: Shadows and eclipses show that light goes straight.

Teacher's Note:
1. Explain each with a line of reasoning.
2. Draw the Sun-Earth-Moon line for eclipses.

Question 2. How can laws of reflection be experimentally verified?
Answer:
1. Spread a white sheet on a table, draw a line MM' and fix a plane mirror upright along it.
2. Fix two pins P and Q in front of the mirror.
3. View the images of P and Q in the mirror and fix two more pins P' and Q' so that they appear in line with the images of P and Q.
4. Remove the mirror and pins. Join QP and extend it to meet the mirror line at O; join P'Q' and extend it to O.
5. Draw the normal ON at O.
6. QP, ON and P'Q' are all drawn on the paper, so they lie in one plane (first law).
7. Measure ∠QON and ∠Q'ON; they are equal (second law).

In simple words: Use pins to mark the incident and reflected rays, then measure the angles at the normal.

Teacher's Note:
1. Draw the labelled figure.
2. Mention both laws in the conclusion.

Question 3. Draw the ray diagrams to represent the formation of image when the object is at infinity and at any distance between infinity and pole of a convex mirror.
Answer:
Object at infinity: parallel rays are reflected so that they appear to come from the principal focus F. The image is formed at F, virtual, erect and highly diminished.

Object anywhere between infinity and the pole: the ray parallel to the axis appears to come from F; the ray directed towards F goes parallel to the axis. The reflected rays diverge and meet when extended backwards, between P and F. The image is virtual, erect and diminished.

In simple words: A convex mirror always gives a small, upright, virtual image between P and F.

Teacher's Note:
1. Use dotted lines for the virtual rays.
2. Label P, F, C, object and image.

Question 4. Describe few sources of light briefly.
Answer:
Natural sources: the Sun and other stars (hot, luminous), and glow-worms, fireflies and some fish (cold sources, bioluminescence).
Artificial sources: a burning candle, an oil lamp, a hurricane lamp (burns kerosene drawn up the wick), an electric bulb (the filament glows when current passes through it) and tube lights.
Sources may also be point or extended, and hot or cold.
In simple words: Natural sources make their own light; artificial sources are made by people.

Teacher's Note:
1. Divide into natural and man-made.
2. Mention a hot and a cold source.

Question 5. Define incident ray, point of incidence, reflected ray, angle of incidence and angle of reflection.
Answer:
Incident ray (AI): the ray of light that falls on the reflecting surface.
Point of incidence (I): the point where the incident ray strikes the surface.
Reflected ray (IB): the ray that bounces off the surface into the same medium.
Normal (IN): the perpendicular to the surface at the point of incidence.
Angle of incidence (∠AIN): the angle between the incident ray and the normal.
Angle of reflection (∠NIB): the angle between the reflected ray and the normal.

In simple words: Both angles are measured from the normal, never from the mirror.

Teacher's Note:
1. Draw and label the figure.
2. The angles are equal (law of reflection).

Question 6. Describe the working of a 'kaleidoscope' and a 'periscope'.
Answer:
Kaleidoscope: it works on multiple reflection. Three plane mirrors inclined at 60° to each other are fixed inside a cylindrical tube, with a few small pieces of coloured glass at one end. The repeated reflections make many images of the glass pieces, which form beautiful patterns when seen from the other end.
Periscope: it is a tube bent twice at right angles, with a plane mirror at each bend inclined at 45° to the tube. Light from the object enters the upper opening, falls on the first mirror at 45° and turns through 90°, travels along the tube, strikes the second mirror and turns through 90° again, and comes out to the eye. It lets us see things that are not in the direct line of sight.

In simple words: A kaleidoscope uses repeated reflections for patterns; a periscope uses two mirrors to see around obstacles.

Teacher's Note:
1. Mention the angles 60° and 45°.
2. Draw the periscope with the light path.

Teacher's Note:
1. Learn the six positions in order.
2. Real images are always inverted.

Question 8. Two plane mirrors are inclined at an angle 'θ'. A light ray incident on one of the mirrors undergoes multiple reflections. The angle of incidence on the first mirror is 15°. After reflection from the second mirror, if the light ray retraces its path, determine the value of θ.

Answer:
The angle of incidence on the first mirror OA is 15°, so the ray makes 90° − 15° = 75° with OA: ∠OAB = 75°.
The ray retraces its path only if it falls normally on the second mirror OB, so ∠ABO = 90°.
In triangle OAB: ∠AOB + ∠OAB + ∠ABO = 180°, so θ + 75° + 90° = 180°.
θ = 180° − 165° = 15°.
In simple words: The ray must hit the second mirror straight on, which fixes the angle between the mirrors.

Teacher's Note:
1. Use the triangle angle sum.
2. Normal incidence means retracing the path.

Question 9. Two identical mirrors are placed in a hollow cylindrical tube as shown in the figure. A light ray incident on one of the mirrors undergoes multiple reflections. If the length of each mirror is 3 m, and the velocity of light is 3 × 10⁸ m s⁻¹, then find the time taken by the light ray incident on the mirror to come out of the hollow cylinder.

Answer:
Each time the ray goes from one mirror to the other it moves ½ m along the tube and 1 m across. In triangle ABC: \( AC = \sqrt{1^2 + (1/2)^2} = \dfrac{\sqrt{5}}{2} \) m.
The tube is 3 m long and each reflection covers ½ m, so the ray makes 3 ÷ ½ = 6 such stretches before it leaves at M.

Total path d = 6 × \( \dfrac{\sqrt{5}}{2} \) = 3√5 m.
Time \( t = \dfrac{d}{c} = \dfrac{3\sqrt{5}}{3 \times 10^{8}} \) = √5 × 10⁻⁸ s (about 2.24 × 10⁻⁸ s).
In simple words: Count six slanting stretches, add their lengths and divide by the speed of light.

Teacher's Note:
1. Use Pythagoras for one stretch.
2. Time = distance ÷ speed.

Question 10. A student constructed a periscope by joining different periscopes P, Q, R, S and T as shown in the figure. Then, (i) how many times does the incident light ray undergo reflection? (ii) when he observes the image of an object using only one periscope, what kind of variation can be observed in the image, when compared with the image formed by the periscope shown in the figure?

Answer:
(i) Each periscope has two mirrors, and the light is reflected once at each mirror. With five periscopes the light undergoes 5 × 2 = 10 reflections.
(ii) With a single periscope the light is reflected only twice, so little light is lost and the image is brighter (clearer). The long chain of ten reflections makes the final image dimmer.
In simple words: Every mirror reflection loses a little light, so fewer reflections give a brighter image.

Teacher's Note:
1. Count the mirrors, not the tubes.
2. Each mirror reflects the ray once.

Question 11. In the figure, AB is an object and A¹B¹ is the image formed when a spherical mirror is placed at point P. Identify the type of mirror and describe the nature of the image. (Where F and C are focus and centre of curvature of the mirror)

Answer:
Here the object lies between F and C. Only a concave mirror forms an inverted image of a real object, and the image is beyond C.

The mirror is a concave mirror. The image is real, inverted and magnified, formed beyond C.
In simple words: An upside-down, bigger image beyond C comes only from a concave mirror.

Teacher's Note:
1. Inverted images are real.
2. Magnified because the object is between C and F.

Question 12. A LASER source placed between two concave mirrors sends light rays as shown in the figure. If PQ and RS are parallel to the principal axis, determine the focal lengths and radius of curvatures of both the mirrors.

Answer:
The ray from the source reaches the first mirror and after reflection goes parallel to the principal axis (towards the second mirror). So the source is at the principal focus of the first mirror.
The parallel ray, after reflection from the second mirror, passes through the laser, so the laser is also at the principal focus of the second mirror.
The source is 10 cm from the first mirror, so f = 10 cm for both mirrors.
Radius of curvature R = 2f = 2 × 10 = 20 cm for both mirrors.
In simple words: The laser sits at the focus of both mirrors, 10 cm from each.

Teacher's Note:
1. Parallel after reflection means the source is at F.
2. Use R = 2f.