CBSE Class 10 Physics Light Reflection and Refraction MCQs Set 14

Answer: (c) (A) and (B)

Explanation:
1. A real image is formed where light rays actually meet, i.e. actually converge at a point.
2. Statements (A) and (B) describe this.
3. (C) and (D) describe a virtual image.

Teacher's Note:
1. Real image: rays actually meet; it can be caught on a screen.
2. Virtual image: rays only appear to meet.

Answer: (c) a horse of less brightness

Explanation:
1. Every part of the lens forms a complete image of the object.
2. Blocking a few strips only reduces the amount of light reaching the image.
3. So a complete horse is seen, but with less brightness.

Teacher's Note:
1. Covering part of a lens never removes part of the image.
2. It only reduces the intensity of the image.

Answer: (c) At infinity

Explanation:
1. Rays from an object at infinity are parallel to the principal axis.
2. A convex lens makes them converge at the focus, forming a real, inverted, point-sized image.

Teacher's Note:
1. Object at infinity gives image at F₂.
2. The image is highly diminished and point-sized.

Answer: (b) focus

Explanation:
1. Sun's rays are practically parallel to the principal axis.
2. A convex lens converges them to the principal focus.

Teacher's Note:
1. Parallel rays meet at the focus after refraction by a convex lens.
2. The distance of this point from the lens is the focal length.

Answer: (b) towards the lens

Explanation:
1. A concave lens always forms a virtual image between the object and the lens, within f.
2. As the object comes closer, the image also comes closer to the lens, with its size increasing.

Teacher's Note:
1. For a concave lens the image is always between O and F₁.
2. Both object and image approach the lens together.

Answer: (b) At twice the focal length

Explanation:
1. A convex lens gives a real, inverted image of the same size when the object is at 2F₁.
2. The image is then also at 2F₂.

Teacher's Note:
1. At 2F both object and image distances are 2f and m = −1.
2. This is how focal length is found in the lab.

Answer: (a) both concave.

Explanation:
1. For a mirror, a negative focal length means a concave mirror (convex mirrors have f positive).
2. For a lens, a negative focal length means a concave (diverging) lens.
3. So both are concave.

Teacher's Note:
1. Concave mirror: f negative. Concave lens: f negative.
2. Convex mirror and convex lens: f positive.

Answer: (d) − f

Explanation:
1. u = −f/2.
2. 1/v = 1/f + 1/u = 1/f − 2/f = −1/f.
3. v = −f; the image is virtual, on the same side as the object, at distance f from the lens.

Teacher's Note:
1. u < f for a convex lens gives a virtual, enlarged image.
2. The negative v shows the image is on the object side.

Answer: (d) A diverging lens always forms a virtual image of a real object.

Explanation:
1. For a concave lens 1/v = 1/f + 1/u with f negative, v is always negative for any real object.
2. So the image is always virtual, whatever the value of u.

Teacher's Note:
1. A diverging lens always gives a virtual, erect, diminished image.
2. No special position is needed.

Answer: (c) increases

Explanation:
1. The bending of light depends on the refractive index of the lens relative to its surroundings.
2. In water the relative refractive index of glass falls (1.5/1.33 instead of 1.5), so the lens bends light less.
3. Its focal length therefore increases.

Teacher's Note:
1. In a denser surrounding medium a lens becomes weaker.
2. If the lens and the liquid had equal refractive indices it would disappear.

Answer: (c) Both A and B are true

Explanation:
1. (A) is true: virtual images are erect (m positive) and real images are inverted (m negative).
2. (B) is also true: concave lens and convex mirror give only virtual, erect images, whereas convex lens and concave mirror give real or virtual images.
3. So both are true.

Teacher's Note:
1. Sign of m tells the type of image.
2. Concave lens and convex mirror always give positive m.

Answer: (c) −50 cm

Explanation:
1. f = 1/P = 1/(−2) = −0.5 m = −50 cm.
2. A negative power means a concave lens.

Teacher's Note:
1. The sign of power and focal length is the same.
2. Convert metres to centimetres at the end.

Answer: (c) 50 cm

Explanation:
1. Power of the combination P = P₁ + P₂ = +3 − 1 = +2 D.
2. f = 1/P = 1/2 m = 50 cm.

Teacher's Note:
1. Powers of lenses in contact add algebraically.
2. Convert to focal length only at the end.

Answer: (d) −80

Explanation:
1. P = P₁ + P₂ = 2.50 − 3.75 = −1.25 D.
2. f = 1/P = 1/(−1.25) = −0.8 m = −80 cm.

Teacher's Note:
1. Add the powers with signs.
2. A negative total power means the combination is diverging.

Answer: (b) 16 cm

Explanation:
1. Erect image means m = +3, so v = 3u.
2. 1/v − 1/u = 1/f gives 1/(3u) − 1/u = 1/12, so −2/(3u) = 1/12 and u = −8 cm, v = −24 cm.
3. Distance between object and image = 24 − 8 = 16 cm.

Teacher's Note:
1. Both object and image are on the same side of the lens for a virtual image.
2. Subtract the distances, do not add them.

Answer: (c) 10 cm

Explanation:
1. The image is the same size as the object when the object is at 2f.
2. Image smaller at 21 cm and larger at 19 cm means 2f lies between 19 and 21 cm, i.e. 2f ≈ 20 cm.
3. Therefore f ≈ 10 cm.

Teacher's Note:
1. Size changes from smaller to larger around u = 2f.
2. Do not give 2f as the focal length.

Answer: (a) Yes, because a real image will be formed.

Explanation:
1. u = −25 cm and f = +20 cm.
2. 1/v = 1/20 − 1/25 = 1/100, so v = +100 cm.
3. v is positive, so the image is real (and inverted) and can be obtained on a screen.

Teacher's Note:
1. An object beyond F of a convex lens gives a real image.
2. A screen can show only real images.

Answer: (a) ½ times

Explanation:
1. u = −36 cm and f = +12 cm.
2. 1/v = 1/12 − 1/36 = 2/36, so v = +18 cm.
3. m = v/u = 18/(−36) = −½, i.e. the image is ½ times the size of the object.

Teacher's Note:
1. m = v/u for lenses.
2. The negative sign means the image is inverted and the size is half.

Answer: (c) 5 dioptre

Explanation:
1. u = −60 cm and v = +30 cm.
2. 1/f = 1/v − 1/u = 1/30 + 1/60 = 1/20, so f = 20 cm = 0.2 m.
3. P = 1/f = 5 D.

Teacher's Note:
1. Convert f to metres before finding the power.
2. 1/0.2 = 5.

Answer: (a) + 5 D

Explanation:
1. f = 20 cm = 0.2 m.
2. P = 1/f = 1/0.2 = +5 D.
3. The power of a convex lens is positive.

Teacher's Note:
1. Convex lens: positive power. Concave lens: negative power.
2. Always use metres.

Answer: (b) Concave mirror

Explanation:
1. A plane mirror, convex mirror and concave lens always form virtual images of a real object.
2. A concave mirror can form a real image when the object is beyond its focus.

Teacher's Note:
1. Real images need converging devices: concave mirror and convex lens.
2. Diverging devices only give virtual images.

Answer: (d) All of these factors

Explanation:
1. The focal length of a lens is controlled by how strongly it bends light.
2. That depends on its radii of curvature, the refractive index of its material and the refractive index of the surrounding medium.

Teacher's Note:
1. Unlike mirrors, lenses refract, so the medium matters.
2. Thicker and more curved lenses have shorter focal lengths.

Answer: (d) First increases and then decreases

Explanation:
1. The beam converges to the focus, where the intensity is maximum.
2. After the focus the rays diverge again, so the intensity falls.

Teacher's Note:
1. Light is most concentrated at the focus.
2. Intensity rises up to F and falls beyond it.

Answer: (c) Real, inverted, diminished and in blue shade

Explanation:
1. An image obtained on a screen is real, and a convex lens makes it inverted.
2. It is diminished because the building is far away.
3. The colour of the building does not change, so the image is blue.

Teacher's Note:
1. Colour of the image is the same as the colour of the object.
2. A virtual image cannot be received on a screen.

Answer: (c) Concave mirror of focal length 24 cm

Explanation:
1. The screen is on the same side as the sun, so the light is reflected: the device is a mirror.
2. A real image of the sun on the screen means a concave mirror.
3. The focal length equals the mirror-to-screen distance, 24 cm.

Teacher's Note:
1. Image on the same side as the object means a mirror.
2. A convex mirror cannot form a real image.

Answer: (c) (C)

Explanation:
1. The image of a distant tree is formed at the focus of the lens.
2. The screen must be on the far side of the lens at 42.7 cm + 15 cm = 57.7 cm.

Teacher's Note:
1. The screen goes at a distance f from the lens, on the side opposite to the object.
2. The other positions are on the wrong side or at the wrong distance.

Answer: (a) Most: f = 10 cm; Least: f = 40 cm; Reason: A convex lens of shorter focal length bends the light rays through larger angles

Explanation:
1. Power P = 1/f, so the lens of the shortest focal length (10 cm) has the greatest power and converges light the most.
2. The lens with the longest focal length (40 cm) converges the least.
3. A short focal length means the rays are bent through larger angles.

Teacher's Note:
1. Short focal length means strong convergence.
2. Powers here are 10 D, 5 D, 3.3 D and 2.5 D.

Answer: (d) −80/3 cm

Explanation:
1. The image is real and inverted, so m = −3 and v = +80 cm.
2. m = v/u gives −3 = 80/u, so u = −80/3 cm ≈ −26.7 cm.

Teacher's Note:
1. For lenses m = v/u; for mirrors m = −v/u.
2. The object distance is negative by the sign convention.

Answer: (d) −40 cm

Explanation:
1. The image can be taken on a screen, so it is real and inverted: m = −3, so v = −3u.
2. 1/v − 1/u = 1/f gives −1/(3u) − 1/u = 1/30, so −4/(3u) = 1/30 and u = −40 cm.
3. Check: v = +120 cm and 1/120 + 1/40 = 1/30 ✓.

Teacher's Note:
1. A real image on a screen means m is negative.
2. Use m = v/u first, then the lens formula.

Answer: (c) Both (I) and (III)

Explanation:
1. The sun is far away, so its image is a sharp, bright, small spot: (II) is correct and (I) is incorrect.
2. The paper takes some time to heat up and burn: (IV) is correct and (III) is incorrect.
3. The incorrect observations are (I) and (III).

Teacher's Note:
1. Image of the sun is a point-sized bright spot.
2. Burning needs time because the paper must reach its ignition temperature.

Answer: (c) < 20 cm or between 20 cm and 40 cm

Explanation:
1. A convex lens can form a twice-enlarged image in two ways.
2. Real, inverted image (m = −2): u = −30 cm, which lies between 20 cm and 40 cm.
3. Virtual, erect image (m = +2): u = −10 cm, which is less than 20 cm.

Teacher's Note:
1. Convex lens gives enlarged images for u between f and 2f (real) and for u less than f (virtual).
2. Both cases are possible.

Answer: (d) 30 cm

Explanation:
1. The object is 5 cm beyond the first focus, so u = −(10 + 5) = −15 cm.
2. 1/v = 1/f + 1/u = 1/10 − 1/15 = 1/30, so v = +30 cm.

Teacher's Note:
1. Distance from the first focus plus f gives the distance from the lens.
2. A real image is on the other side of the lens.

Answer: (d) −3.75 dioptre

Explanation:
1. Power of the combination P = 1/0.8 = 1.25 D.
2. Power of the 20 cm lens P₁ = 1/0.2 = 5 D.
3. P₂ = P − P₁ = 1.25 − 5 = −3.75 D.

Teacher's Note:
1. Powers of lenses in contact add up.
2. Subtract the known power from the total to get the unknown one.

Answer: (b) diverging lens

Explanation:
1. In a liquid of higher refractive index than glass, light bends away from the normal on entering the lens.
2. A convex lens then diverges the rays, so it behaves as a diverging lens.

Teacher's Note:
1. The behaviour depends on the refractive index of the lens relative to the medium.
2. If the indices are equal, the lens does not bend light at all.

Answer: (d) red light

Explanation:
1. Red light has the smallest refractive index in glass, so it bends the least.
2. The lens bends it the least and focuses it farthest away, so the focal length is maximum for red light.

Teacher's Note:
1. Violet bends most and has the shortest focal length; red bends least and has the longest.
2. This difference causes chromatic aberration.

Answer: (c) concave lens forms virtual, erect and diminished images irrespective of the position of the object.

Explanation:
1. A concave lens always forms a virtual, erect and diminished image.
2. So its magnification is always positive and less than 1.

Teacher's Note:
1. Virtual, erect, diminished: 0 < m < 1.
2. The same holds for a convex mirror.

Answer: (b) Beyond focus

Explanation:
1. A convex lens forms a real image whenever the object is beyond the focus (u > f).
2. Within the focus the image is virtual.

Teacher's Note:
1. Real image: object beyond F. Virtual image: object between O and F.
2. Do not restrict it to 2F.

Answer: (c) A is true, but R is false.

Explanation:
1. A virtual image is formed where rays only appear to meet, so it cannot be received on a screen.
2. Assertion is true.
3. Reason is false because real images, not virtual ones, are formed by the actual meeting of rays.

Teacher's Note:
1. Real image: rays actually meet, can be projected.
2. Virtual image: rays appear to meet, cannot be projected.

Answer: (d) A is false, but R is true.

Explanation:
1. A convex lens can also give diverging rays (for example when the object is within its focal length), so the lens need not be convex and A is false.
2. A concave lens always makes rays diverge, so R is true.

Teacher's Note:
1. Diverging rays do not prove the lens is concave or convex.
2. Check the position of the source for a convex lens.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. P = 1/f, so power is inversely proportional to focal length: A is true.
2. The SI unit of power is dioptre: R is true.
3. But the unit does not explain the inverse relation.

Teacher's Note:
1. A unit is a label, not a reason for a relation.
2. P = 1/f with f in metres.