CBSE Class 10 Science Light - Reflection and Refraction Important Questions Set 01

Answer: n(glass/water) = n(glass) ÷ n(water) = \( \dfrac{3/2}{4/3} = \dfrac{9}{8} \) = 1.125.

Teacher's Note:
1. Divide the absolute refractive index of the second medium by that of the first.
2. Both values are with respect to air, so they can be compared directly.

Answer: f = −200 cm = −2 m, so P = 1/f = \( \dfrac{1}{-2} \) = −0.5 D.

Teacher's Note:
1. A concave lens has a negative focal length and so negative power.
2. Always convert the focal length to metres before using P = 1/f.

Answer: f = R/2 = 20/2 = 10 cm.

Teacher's Note:
1. The focal length is half the radius of curvature.
2. The sign depends on the type of mirror: negative for concave and positive for convex.

Answer: 0°. A normal ray has an angle of incidence of 0°, and the angle of reflection equals the angle of incidence.

Teacher's Note:
1. The ray retraces its own path.
2. Angles are always measured from the normal, not from the mirror surface.

Answer: +1. The image is virtual, erect and of the same size as the object.

Teacher's Note:
1. The positive sign shows that the image is erect.
2. A magnification of magnitude 1 means that the image is the same size as the object.

Answer: The image is virtual, erect and enlarged (three times the object size).

Teacher's Note:
1. A positive magnification means that the image is erect and virtual.
2. A concave mirror forms such an image only when the object is between the pole and the focus.

Answer: v = c/n = \( \dfrac{3\times10^{8}}{1.5} \) = 2 × 10⁸ m/s.

Teacher's Note:
1. The speed of light is smaller in a denser medium.
2. Use n = c/v for the absolute refractive index.

Answer: (b) f positive, u negative. The focal length of a convex mirror is positive and the object is placed on the left of the mirror, so u is negative.

Teacher's Note:
1. Follow the Cartesian sign convention with the pole as the origin.
2. Distances measured against the direction of the incident light are negative.

Answer: (c) between P and F. In this position the image is behind the mirror, erect and magnified.

Teacher's Note:
1. This is the situation of a shaving mirror.
2. Between F and C or beyond C the image is real and inverted.

Answer: (c) aperture. Mirror formulae are valid for mirrors of small aperture.

Teacher's Note:
1. The centre of curvature and the principal focus are points, not lengths.
2. The relation R = 2f is a formula, not a name.

Answer: (d) both (b) and (c). For a lens m = v/u = h′/h.

Teacher's Note:
1. The negative sign in m = −v/u belongs to the mirror formula.
2. The ratio of heights and the ratio of distances give the same value.

Answer: (a) F₂. A convex lens converges a beam parallel to its axis at the principal focus on the other side of the lens.

Teacher's Note:
1. This is why the lens is called a converging lens.
2. A ray passing through the optical centre goes straight.

Answer: (c). A is true because light is 1.33 times slower in water than in vacuum. R is false because water is optically denser than air.

Teacher's Note:
1. A larger refractive index means a more optically dense medium.
2. Water has a larger refractive index than air.

Answer: (a). Both are true. The focus of a concave mirror lies in front of it, against the incident light, so f is negative, and the focus of a convex mirror lies behind it, so f is positive.

Teacher's Note:
1. The incident light is taken as travelling left to right, with the pole at the origin.
2. The convention explains the sign of f.

Answer: Real image, so it is inverted: h = +2 cm, h′ = −3 cm, u = −15 cm.
m = h′/h = −3/2 = −v/u, so v = 1.5u = 1.5 × (−15) = −22.5 cm.
The image is 22.5 cm in front of the mirror.

Teacher's Note:
1. Use m = h′/h = −v/u to find v directly.
2. The negative sign of v shows that the image is on the same side as the object.

Answer: f = 1/P = −0.25 m = −25 cm. The lens is a concave (diverging) lens.
The image formed is virtual and erect, and it is smaller than the object.

Teacher's Note:
1. A negative power always means a diverging lens.
2. A concave lens forms the same type of image for every position of the object.

Answer: 1. A pencil kept partly in a glass of water appears bent at the water surface.
2. A coin in a bucket of water appears raised, and water in a tank looks shallower than its real depth.

Teacher's Note:
1. Refraction occurs because light changes speed when it passes from one medium to another.
2. Twinkling of stars is also a result of refraction.

Answer: A real image is formed when the reflected or refracted rays actually meet. It can be obtained on a screen and is always inverted.
A virtual image is formed when the rays only appear to meet. It cannot be obtained on a screen and is erect.

Teacher's Note:
1. Plane mirrors always form virtual images.
2. Concave mirrors form both real and virtual images.

Answer: (a) A concave mirror is used in headlights because with the bulb at its focus it sends out a strong parallel beam of light.
(b) A convex mirror is used as the rear-view mirror because it forms an erect image and gives a wide field of view.

Teacher's Note:
1. Concave mirrors converge light, while convex mirrors diverge it.
2. A convex mirror always forms a diminished image.

Answer: u = −10 cm, f = +15 cm.
\( \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{15} + \dfrac{1}{10} = \dfrac{5}{30} \), so v = +6 cm.
The image is 6 cm behind the mirror, virtual, erect and diminished (m = −v/u = 0.6).

Teacher's Note:
1. Use the mirror formula with correct signs for a convex mirror.
2. A positive v shows that the image is behind the mirror.

Answer: n(turpentine/alcohol) = 1.47 ÷ 1.36 ≈ 1.08.
Light travels faster in the medium with the smaller refractive index, so its speed is more in alcohol.

Teacher's Note:
1. Refractive index and speed of light in a medium are inversely related.
2. The relative refractive index is greater than 1, so turpentine oil is denser than alcohol.

Answer: (a) The power of a lens is the reciprocal of its focal length in metres, P = 1/f. Its SI unit is the dioptre (D).
(b) The image is the same size, so the needle is at 2F, and u = −v = −50 cm. The needle is placed 50 cm from the lens.
2f = 50 cm, so f = 25 cm = 0.25 m and P = 1/0.25 = +4 D.

Teacher's Note:
1. For a convex lens, an equal-sized real image forms when the object is at 2F.
2. In that case, object and image are at equal distances from the lens.

Answer: u = −16 cm, h = 2 cm, h′ = −3 cm.
(a) m = h′/h = −1.5 = −v/u, so v = 1.5u = −24 cm. The image is 24 cm in front of the mirror.
(b) \( \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} = -\dfrac{1}{24} - \dfrac{1}{16} = -\dfrac{5}{48} \), so f = −9.6 cm.

Teacher's Note:
1. Find v from the magnification and then f from the mirror formula.
2. Negative values of v and f are expected for a real image formed by a concave mirror.

Answer: n = c/v = \( \dfrac{3\times10^{8}}{1.24\times10^{8}} \) ≈ 2.42.
The refractive index of diamond is about 2.42, which is very high, so it is optically very dense.

Teacher's Note:
1. The absolute refractive index has no unit.
2. This high value makes diamond sparkle strongly.

Answer: (a) Convex lens, object between O and F: the image is on the same side as the object, virtual, erect and enlarged.
(b) Convex lens, object at 2F: the image is at 2F on the other side, real, inverted and the same size as the object.
(c) Concave lens, object anywhere in front of it: the image is between the optical centre and F on the same side as the object, virtual, erect and diminished.

Teacher's Note:
1. Practise drawing the rays through the optical centre, parallel to the axis and through the focus.
2. A magnifying glass uses the case described in part (a).

Answer: u = −15 cm, f = +20 cm, h = 1 cm.
\( \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{20} - \dfrac{1}{15} = \dfrac{3-4}{60} = -\dfrac{1}{60} \), so v = −60 cm.
m = v/u = (−60)/(−15) = +4, so h′ = 4 cm.
The image is virtual and erect, enlarged to 4 cm, formed 60 cm from the lens on the same side as the object.

Teacher's Note:
1. The object is within the focal length, so a virtual image is expected.
2. A positive magnification confirms an erect image.

Answer: u = −15 cm, f = −10 cm, h = 1 cm.
\( \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = -\dfrac{1}{10} + \dfrac{1}{15} = -\dfrac{1}{30} \), so v = −30 cm.
m = −v/u = −(−30)/(−15) = −2, so h′ = −2 cm.
The image is real, inverted and enlarged, 2 cm in size, formed 30 cm in front of the mirror.

Teacher's Note:
1. The object lies between F and C, so the image is beyond C.
2. A negative magnification shows that the image is inverted.

Answer: (a) At infinity: image at F, real, inverted, highly diminished.
(b) Beyond C: image between F and C, real, inverted, diminished.
(c) At C: image at C, real, inverted, same size.
(d) Between C and F: image beyond C, real, inverted, enlarged.
(e) At F: image at infinity, real, inverted, highly enlarged.
(f) Between F and P: image behind the mirror, virtual, erect, enlarged.

Teacher's Note:
1. This table is the base for most mirror questions.
2. Only the last position gives a virtual image.