CBSE Class 10 Science Light and the Human Eye Important Questions Set 01

Answer: Power of the combination = power of convex lens + power of concave lens, so \( \dfrac{1}{F} = \dfrac{1}{f} + \dfrac{1}{(-f)} = 0 \). Hence F is infinite and the combination behaves like a plane glass sheet.

Teacher's Note:
1. Powers of lenses in contact add algebraically.
2. A concave lens has negative focal length and power, which cancels the convex lens.

Answer: Light travels fastest in water. Speed in a medium is v = c/n, so the medium with the smallest refractive index (water, 1.33) has the largest speed.

Teacher's Note:
1. Speed of light in a medium is inversely proportional to its refractive index.
2. Compare the given indices and pick the smallest.

Answer: Power P = 1/f, with f in metres, so power is the reciprocal of focal length. The lens of focal length 20 cm has power +5 D against +2.5 D for the 40 cm lens, so the 20 cm lens bends light more and converges it more.

Teacher's Note:
1. A shorter focal length means a greater power and stronger bending of rays.
2. Convert the focal lengths to metres before finding the powers.

Answer: At the first face the ray bends towards the normal and at the second face it bends away from the normal by the same angle, because the two faces are parallel. The two bendings cancel, so the emergent ray is parallel to the incident ray, though shifted sideways (lateral displacement).

Teacher's Note:
1. The angle of refraction at the first face equals the angle of incidence at the second face.
2. This is why a slab does not change the direction of light, only its position.

Answer: 1. Only a concave mirror forms a real image on a screen, and the negative magnification shows the image is real and inverted. So it is a concave mirror.
2. m = −v/u = −1 gives v = u = −50 cm, so the object and the image both lie at the centre of curvature, and the distance between them is zero.
3. R = 50 cm, so f = R/2 = −25 cm. Check: \( \dfrac{1}{f} = \dfrac{1}{-50} + \dfrac{1}{-50} = -\dfrac{1}{25} \).

Teacher's Note:
1. A real image of the same size is formed only when the object is at the centre of curvature.
2. Use m = −v/u and then the mirror formula to check the focal length.

Answer: 1. The object must lie between the pole and the focus, that is within 0 to 15 cm from the mirror.
2. The image is virtual, erect and larger than the object (magnified).
3. At 30 cm the object is at the centre of curvature (R = 2f = 30 cm). The image is real, inverted, of the same size, and formed at 30 cm in front of the mirror.

Teacher's Note:
1. A concave mirror gives an erect image only when the object is closer than its focus.
2. Remember that R = 2f.

Answer: The absolute refractive index of a medium is n = c/v, where c is the speed of light in vacuum and v is the speed in the medium. The refractive index of medium 2 with respect to medium 1 is \( n_{21} = \dfrac{n_2}{n_1} = \dfrac{c/v_2}{c/v_1} = \dfrac{v_1}{v_2} \), the ratio of the speed of light in medium 1 to that in medium 2.

Teacher's Note:
1. Write each absolute index as c divided by the speed in that medium.
2. The factor c cancels in the ratio.

Answer: The line joining the centre of curvature to any point of the mirror is a radius, and so it is normal to the surface at that point. A ray travelling along it strikes the mirror at an angle of incidence 0°, so the angle of reflection is also 0° and the ray retraces its path.

Teacher's Note:
1. The radius of a spherical surface is always perpendicular to the surface.
2. A ray along the normal is not deviated on reflection.

Answer: (a) Concave mirror as well as convex lens. A point source at the focus of a concave mirror or of a convex lens gives a parallel beam after reflection or refraction.

Teacher's Note:
1. Rays from the focus become parallel after passing through a convex lens or reflecting from a concave mirror.
2. Convex mirrors and concave lenses only diverge light.

Answer: (b) −20 cm. The image is real and inverted, so m = −5/10 = −0.5 = −v/u, giving u = −60 cm (with v = −30 cm); then \( \dfrac{1}{f} = \dfrac{1}{-30} + \dfrac{1}{-60} = -\dfrac{1}{20} \), so f = −20 cm.

Teacher's Note:
1. Find the object distance from the magnification first.
2. Then apply the mirror formula with the sign convention.

Answer: \( n_{dg} = \dfrac{n_d}{n_g} \), so \( n_d = n_{dg} \times n_g = 1.6 \times 1.5 = 2.4 \).

Teacher's Note:
1. The refractive index with respect to glass is the ratio of the two absolute indices.
2. The absolute index of diamond is therefore 2.4.

Answer: (a) light is least scattered. At noon the Sun is overhead and its light travels the shortest path through the atmosphere, so very little blue and violet light is scattered away.

Teacher's Note:
1. A shorter path through the air means less scattering.
2. All colours then reach the eye in nearly the right proportion.

Answer: This is due to atmospheric refraction. The density of air, and so its refractive index, changes with height, so rays from the top and bottom edges of the Sun are refracted by different amounts and the disc looks flattened vertically. Rays from the side edges are refracted equally, so the Sun keeps its width.

Teacher's Note:
1. Atmospheric refraction is stronger near the horizon.
2. Different layers of air bend rays from the top and bottom of the Sun by different amounts.

Answer: At noon sunlight passes through a relatively thin layer of the atmosphere to reach us. Only a small part of the blue light is scattered, so all colours reach the eye almost in the original proportion and the Sun appears white.

Teacher's Note:
1. Scattering depends on the length of the path through air.
2. A short path means little loss of blue light.

Answer: A concave (diverging) lens is needed. It must form the image of a distant object at the far point, so u = −∞ and v = −120 cm. Then \( \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} = -\dfrac{1}{120} \), so f = −120 cm and \( P = \dfrac{100}{-120} \approx -0.83 \) D.

Teacher's Note:
1. The corrective lens must bring the image of a far object to the person's far point.
2. The focal length equals the far point distance with a negative sign.

Answer: The two refracting faces of a prism are inclined to each other, so different colours are bent by different amounts and spread out as a spectrum. The faces of a slab are parallel, so every colour bends at the first face and bends back at the second, and the colours emerge together as white light parallel to the incident beam.

Teacher's Note:
1. Dispersion needs the colours to be separated and not recombined.
2. In a slab the second face reverses the bending made by the first.

Answer: A concave lens is required. Here u = −∞ and v = −80 cm, so \( \dfrac{1}{f} = \dfrac{1}{-80} - \dfrac{1}{-\infty} = -\dfrac{1}{80} \) and f = −80 cm. Power \( P = \dfrac{100}{-80} = -1.25 \) D.

Teacher's Note:
1. Take the focal length equal to the far point distance and negative.
2. Power in dioptres is 100 divided by f in centimetres.

Answer: A convex lens is needed. It must form a virtual image at 100 cm of an object placed at 25 cm, so u = −25 cm and v = −100 cm. Then \( \dfrac{1}{f} = \dfrac{1}{-100} - \dfrac{1}{-25} = \dfrac{3}{100} \), so f = 100/3 cm ≈ 33.3 cm and P = +3 D.

Teacher's Note:
1. The lens must make an object at 25 cm appear to be at the defective near point.
2. A positive power confirms a convex lens.

Answer: Starlight passes through layers of air whose density and refractive index keep changing slightly. The path of the light therefore keeps shifting, so the apparent position and the brightness of the star change from moment to moment, and it seems to twinkle.

Teacher's Note:
1. Twinkling is caused by atmospheric refraction of starlight.
2. Stars act as point sources, so small changes in the path are easily noticed.

Answer: Planets are much closer to us than stars and so act as extended sources made of many point sources. The variations in light from the individual points average out, so the total brightness stays almost steady and no twinkling is seen.

Teacher's Note:
1. A planet is an extended source of light and not a point source.
2. Changes from different points cancel out.

Answer: Near the horizon sunlight has to travel through a thicker layer of air. Most of the blue light is scattered away on the way, and the red light, which is scattered the least, reaches the eye, so the Sun looks red.

Teacher's Note:
1. The path through the atmosphere is longest at sunrise and sunset.
2. Light of longer wavelength is scattered less.

Answer: The molecules of air scatter blue light of short wavelength much more than red light, so scattered blue light reaches our eyes from all directions. In space there is no atmosphere to scatter light, so the sky appears dark.

Teacher's Note:
1. Blue light is scattered more strongly than red light.
2. Without air there is nothing to scatter sunlight.