CBSE Class 7 Science Advanced Physics Light MCQs Set 04

Answer: (b) 60°

Explanation:
1. The angle between the incident and the reflected rays is ∠i + ∠r = 2∠i = 60°, so ∠i = 30°.
2. Glancing angle = 90° − 30° = 60°.

Teacher's Note:
1. Angle between the rays = 2i.
2. Glancing angle = 90° − i.

Answer: (d) All of these

Explanation:
1. For a plane mirror, d = 180° − 2i.
2. Since i = r, d = 180° − 2r.
3. Also i = 90° − α, so d = 180° − 2(90° − α) = 2α.
4. All three are equal: (d).

Teacher's Note:
1. α is the glancing angle: α = 90° − i.
2. i = r, so (a) and (b) are the same.

Answer: (b) A diminished virtual image

Explanation:
1. A convex mirror always forms an erect, virtual and diminished image whatever the position of the object.
2. So (b).

Teacher's Note:
1. Convex mirrors never form real images.
2. The image is smaller than the object.

Answer: (c) 45°

Explanation:
1. The ray makes 45° with the vertical mirror (45° from the mirror surface), so it makes 90° − 45° = 45° with the normal there, and it travels at 45° down.
2. At the horizontal mirror it also makes 45° with the surface, so the angle from the normal is 90° − 45° = 45°.
3. Hence i′ = 45°.

Teacher's Note:
1. Angle with the mirror plus angle with the normal = 90°.
2. With perpendicular mirrors, the angles stay at 45° each time.

Answer: (b) 3 m s⁻¹

Explanation:
1. In a plane mirror the image is as far behind the mirror as the object is in front.
2. So when the object moves 3 m towards the mirror every second, the image also moves 3 m towards the mirror every second: 3 m s⁻¹.
3. The speed of the image equals the speed of the object, but the directions are opposite.

Teacher's Note:
1. Image distance = object distance at every moment.
2. 6 m s⁻¹ is the speed of the image relative to the object, not relative to the mirror.

Answer: (d) All of these

Explanation:
1. Shadows and eclipses form because light travels in straight lines.
2. The direction of light is not changed in any of them: (d).

Teacher's Note:
1. Rectilinear propagation of light.
2. Eclipses are large shadows.

Answer: (b) 683

Explanation:
1. One watt of light power at the most sensitive wavelength is equal to 683 lumen.
2. Answer (b).

Teacher's Note:
1. Remember 683, not 638 or 863.
2. 12.56 is used for the candle power and lumen relation.

Answer: (c) Virtual and erect

Explanation:
1. A plane mirror forms an image behind the mirror that cannot be caught on a screen, so it is virtual.
2. It is the same way up as the object, so it is erect: (c).

Teacher's Note:
1. Plane mirror image: virtual, erect, same size, laterally inverted.
2. Left and right are interchanged, but it is not upside down.

Answer: (b) Concave mirror

Explanation:
1. A rear-view mirror is a convex mirror, and a convex mirror gives a diminished image.
2. A plane mirror gives an image of the same size.
3. A concave mirror gives a magnified virtual image when the object is between the pole and the focus: (b).

Teacher's Note:
1. Shaving mirrors are concave.
2. Object between P and F: magnified, virtual, erect.

Answer: (a) ACB

Explanation:
1. Light enters the drop and splits into colours (A).
2. The coloured rays are reflected from the back of the drop (C).
3. The reflected rays refract again as they leave the drop (B). Order: A, C, B: (a).

Teacher's Note:
1. Refraction and dispersion, reflection, refraction again.
2. A rainbow is seen after this.

Answer: (a) A → r; B → p; C → s; D → q

Explanation:
1. Pole to centre of curvature is the radius of curvature (A → r).
2. The focal length of a convex mirror is R/2 (B → p).
3. Dispersion of light gives a rainbow (C → s).
4. The focal length of the human eye lens is about 2.5 cm (D → q).
5. Answer (a).

Teacher's Note:
1. Focal length = R/2.
2. The eye lens focal length is the book's value of 2.5 cm.

Answer: (b) 3

Explanation:
1. θ = 90°, so 360°/θ = 4, which is even.
2. n = 4 − 1 = 3.

Teacher's Note:
1. n = 360°/θ − 1.
2. Perpendicular mirrors: 3 images.

Answer: (a) Greater than 15 cm

Explanation:
1. f = 10 cm, so C is at 20 cm. The object at 15 cm lies between F and C.
2. For an object between F and C the image forms beyond C, so it is farther than the object.
3. Using 1/v = 1/f − 1/u (magnitudes) gives v = 30 cm, which is greater than 15 cm: (a).

Teacher's Note:
1. Object between F and C: image beyond C, real, inverted, magnified.
2. Quick check: 1/10 − 1/15 = 1/30.

Answer: (c) 1 : 2

Explanation:
1. f(convex) = 10/2 = 5 cm and f(concave) = 20/2 = 10 cm.
2. Ratio = 5 : 10 = 1 : 2.

Teacher's Note:
1. f = R/2.
2. Ratio of focal lengths = ratio of radii = 10 : 20.

Answer: (d)

Explanation:
1. (a) Concave mirror: a ray parallel to the axis goes through F: correct.
2. (b) Concave mirror: a ray directed through F is reflected parallel to the axis: correct.
3. (c) Convex mirror: a ray directed towards F is reflected parallel to the axis: correct.
4. (d) Convex mirror: the ray is directed towards C, so it must be reflected back along its own path, but it is drawn reflected parallel to the axis: incorrect.

Teacher's Note:
1. Concave: parallel ray through F; ray through F parallel.
2. Convex: ray towards F parallel; ray towards C retraces.

Answer: (b) 60°

Explanation:
1. The reflected ray makes 30° with the mirror, so ∠r = 90° − 30° = 60° and ∠i = 60°.
2. Angle of deviation = 180° − 2 × 60° = 60°.

Teacher's Note:
1. Angle with the mirror: 90° minus the angle with the normal.
2. d = 180° − 2i.

Answer: (d) i = (180 – d)/2

Explanation:
1. d = 180° − 2i, so 2i = 180° − d and i = (180° − d)/2: (d) is correct.
2. (a) d = 2i is wrong, (b) d = a + i + r is wrong, and (c) is wrong because d = 2a, not 180° − 2a.

Teacher's Note:
1. i = r and a = 90° − i.
2. d = 180° − 2i = 2a.

Answer: (d) All of these

Explanation:
1. Object between P and F: magnified virtual image.
2. Object beyond C: diminished real image.
3. Object between F and C: magnified real image.
4. So all of these are possible: (d).

Teacher's Note:
1. Concave mirror images depend on the object position.
2. The image size and nature change as the object moves.

Answer: (a) 90°

Explanation:
1. The ray makes 45° with the vertical mirror, so with the horizontal normal i = r = 45°.
2. At the second mirror i′ = r′ = 45°.
3. 2i + i′ − r′ = 90° + 45° − 45° = 90°.

Teacher's Note:
1. Angle of incidence equals angle of reflection at each mirror.
2. i′ and r′ cancel each other.

Answer: (b) Moves towards

Explanation:
1. The image is always as far behind the mirror as the object is in front.
2. When the mirror moves towards the object, the distance of the object from the mirror becomes smaller, so the image distance behind the mirror also becomes smaller.
3. The image moves towards the mirror: (b).

Teacher's Note:
1. Image distance = object distance.
2. A smaller object distance means a smaller image distance.