CBSE Class 10 Physics Light Reflection and Refraction MCQs Set 15

Answer: (b) The point on the principal axis where rays parallel to the axis actually meet after reflection

Explanation:
1. Rays travelling parallel to the principal axis converge to the principal focus after reflecting from a concave mirror, so the focus is real.

Teacher's Note:
1. A convex mirror has a virtual focus behind it.

Answer: (a) 10 cm

Explanation:
1. Focal length f = R/2 = 20/2 = 10 cm (negative for a concave mirror, positive for a convex mirror).

Teacher's Note:
1. The focus lies midway between the pole and the centre of curvature.

Answer: (d) Concave mirror

Explanation:
1. A concave mirror forms an erect, enlarged, virtual image when the object is between the pole and the focus.

Teacher's Note:
1. Convex mirrors always diminish.

Answer: (c) always forms an erect, diminished image and gives a wide field of view

Explanation:
1. Being curved outwards, it spreads the reflected rays and lets the driver see a larger area, although the images are smaller.

Teacher's Note:
1. Plane mirrors give a narrower field of view.

Answer: (b) +16 cm

Explanation:
1. f = R/2 = 32/2 = 16 cm, and it is taken positive for a convex mirror.

Teacher's Note:
1. The focus lies behind the mirror.

Answer: (a) 30 cm in front of the mirror

Explanation:
1. For a real image m = -3, so v = -m u = -(-3)(-10) = -30 cm: the image is 30 cm in front of the mirror.

Teacher's Note:
1. Negative v means the image is on the object side.

Answer: (d) between the pole and the principal focus

Explanation:
1. Only an object placed within the focal length gives a virtual, erect, magnified image in a concave mirror.

Teacher's Note:
1. This is the principle of shaving and make-up mirrors.

Answer: (c) either plane or convex

Explanation:
1. Plane mirrors and convex mirrors always form virtual erect images; a concave mirror gives inverted real images for far objects.

Teacher's Note:
1. Convex gives diminished, plane gives same size.

Answer: (b) 0 to 15 cm

Explanation:
1. The object must lie between the pole and the focus, i.e. within 15 cm. The image is virtual and larger.

Teacher's Note:
1. Beyond the focus the image is real and inverted.

Answer: (a) Concave mirror, as the bulb at its focus gives a strong, nearly parallel beam

Explanation:
1. A source at the principal focus of a concave mirror sends out parallel reflected rays.

Teacher's Note:
1. The same idea is used in torches.

Answer: (d) Concave mirror, because it converges parallel sunlight to a point

Explanation:
1. A large concave reflector concentrates sunlight at its focus, producing high temperature there.

Teacher's Note:
1. The furnace is placed at the focus.

Answer: (c) 6 cm behind the mirror

Explanation:
1. 1/v = 1/f - 1/u = 1/15 + 1/10 = 1/6, so v = +6 cm. The image is virtual, erect and diminished.

Teacher's Note:
1. m = -v/u = +0.6.

Answer: (b) virtual, erect and of the same size as the object

Explanation:
1. A positive sign means a virtual erect image and the magnitude 1 means equal size.

Teacher's Note:
1. Real images have negative magnification.

Answer: (a) 2.14 cm

Explanation:
1. f = +15 cm; 1/v = 1/15 + 1/20 = 7/60, so v = 60/7 = 8.57 cm. m = -v/u = 3/7, so image height = 5 x 3/7 = 2.14 cm (virtual, erect).

Teacher's Note:
1. 8.57 cm is the image distance, not the height.

Answer: (d) 54 cm from the mirror

Explanation:
1. 1/v = 1/f - 1/u = -1/18 + 1/27 = -1/54, so v = -54 cm; m = -2 and the image is real, inverted and 14 cm tall.

Teacher's Note:
1. A real image can be caught on a screen.

Answer: (c) 17.14 cm in front of the mirror

Explanation:
1. f = -12 cm. 1/v = -1/12 + 1/40 = -7/120, so v = -120/7 = -17.14 cm (real, inverted, 1.07 cm tall).

Teacher's Note:
1. The object is beyond C, so the image is between F and C.

Answer: (b) 36.7 cm behind the mirror

Explanation:
1. f = +40 cm; 1/v = 1/40 + 1/450 = 49/1800, so v = +36.73 cm; m = 0.0816, so the image is about 0.139 m tall.

Teacher's Note:
1. Convex mirrors give small virtual images.

Answer: (a) 90 cm (R = -90 cm)

Explanation:
1. v = -m u = -(3)(-30) = +90 cm; 1/f = 1/90 - 1/30 = -1/45, so f = -45 cm and R = 2f = -90 cm.

Teacher's Note:
1. The sign shows a concave mirror.

Answer: (d) -70 cm and +2

Explanation:
1. 1/f = 1/v + 1/u = 1/70 - 1/35 = -1/70, so f = -70 cm; m = -v/u = -70/(-35) = +2.

Teacher's Note:
1. Virtual and magnified image.

Answer: (c) 45 cm

Explanation:
1. For an object at infinity the image forms at the focus, so |f| = 45 cm.

Teacher's Note:
1. This is how a quick focal length is measured.

Answer: (b) 60 cm (real image) or 20 cm (virtual image)

Explanation:
1. Real: v = 2u so 3/(2u) = -1/40, u = -60 cm. Virtual: v = -2u so 1/(2u) = -1/40, u = -20 cm. Both are possible.

Teacher's Note:
1. Magnified images exist for u between F and C (real) and inside F (virtual).

Answer: (a) 10 cm from the mirror

Explanation:
1. v = -m u = -30 cm and 1/f = -1/30 - 1/60 = -1/20, so f = -20 cm. For m = +2, v = -2u: 1/(2u) = -1/20 gives u = -10 cm.

Teacher's Note:
1. Inside the focus for a virtual image.

Answer: (d) 1/v + 1/u = 1/f

Explanation:
1. With the new Cartesian sign convention, 1/v + 1/u = 1/f for spherical mirrors.

Teacher's Note:
1. For lenses, 1/v - 1/u = 1/f.

Answer: (c) retraces its path

Explanation:
1. It strikes the mirror normally, so it is reflected back along the same path.

Teacher's Note:
1. Normal incidence means angle of incidence 0.

Answer: (b) m = h'/h = -v/u

Explanation:
1. With the sign convention, m = h2/h1 = -v/u; negative values indicate real and inverted images.

Teacher's Note:
1. Positive m means virtual and erect.