Exercises
Q1. Define the principal focus of a concave mirror.
Answer:
The principal focus of a concave mirror is a point on its principal axis where rays of light parallel to the principal axis converge after reflection from the mirror.
Q2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer:
Given radius of curvature R = 20 cm. For a spherical mirror, focal length f = R/2 = 20/2 = 10 cm.
Q3. Name a mirror that can give an erect and enlarged image of an object.
Answer:
A concave mirror can give an erect and enlarged image when the object is placed between its pole and principal focus.
Q4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer:
We prefer a convex mirror as a rear-view mirror because it always produces an erect image, and it gives a wider field of view, allowing the driver to see more area behind the vehicle. The image is also diminished, giving a better overall view.
Q5. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
Given radius of curvature R = 32 cm. For a convex mirror, focal length f = R/2 = 32/2 = 16 cm. (Note: Convex mirror has a positive focal length according to sign convention.)
Q6. A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
Answer:
Given magnification m = -3 (real image is inverted), object distance u = -10 cm (u is negative as per sign convention). Magnification m = -v/u, so -3 = -v/(-10) => -3 = v/10 => v = -30 cm. The negative sign indicates the image is on the same side as the object, 30 cm in front of the mirror.
Q7. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
When a ray of light travels from air (rarer medium) to water (denser medium), it slows down. This causes the ray to bend towards the normal at the point of incidence, as per Snell's law.
Q8. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 10? m/s.
Answer:
Given refractive index of glass n = 1.50, speed of light in vacuum c = 3 × 10? m/s. Refractive index n = c/v. Therefore, speed of light in glass v = c/n = (3 × 10?)/1.50 = 2 × 10? m/s.
Q9. Find the power of a concave lens of focal length 2 m.
Answer:
Given focal length f = -2 m (concave lens has negative focal length). Power P = 1/f (in metres) = 1/(-2) = -0.5 D. The power is -0.5 dioptre.
Q10. Which of the following lenses would you prefer to use while reading small letters found in a dictionary? (a) A convex lens of focal length 50 cm. (b) A concave lens of focal length 50 cm. (c) A convex lens of focal length 5 cm. (d) A concave lens of focal length 5 cm.
Answer:
The correct answer is (c) A convex lens of focal length 5 cm. A convex lens acts as a magnifying glass. For reading small letters, we need a magnifying glass that produces a large, erect virtual image. A shorter focal length gives greater magnification. A convex lens of small focal length (5 cm) provides a larger magnification compared to one with 50 cm focal length. Concave lenses always produce diminished images, so they are not suitable.
Q11. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer:
Given: Object height h = 5.0 cm, object distance u = -20 cm (negative as per sign convention), radius of curvature R = 30 cm for convex mirror. Focal length f = R/2 = +15 cm (positive for convex mirror). Mirror formula: 1/v + 1/u = 1/f => 1/v + 1/(-20) = 1/15 => 1/v = 1/15 + 1/20 = (4+3)/60 = 7/60 => v = 60/7 ? 8.57 cm. Positive v indicates image is behind the mirror (virtual). Magnification m = -v/u = -(8.57)/(-20) = 0.4285. Image height h' = m × h = 0.4285 × 5 = 2.14 cm. Image is virtual, erect, and diminished (size 2.14 cm), located 8.57 cm behind the mirror.
Q12. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer:
Given: Object distance u = -10 cm (negative as per sign convention), focal length f = +15 cm (convex mirror). Mirror formula: 1/v + 1/u = 1/f => 1/v + 1/(-10) = 1/15 => 1/v = 1/15 + 1/10 = (2+3)/30 = 5/30 = 1/6 => v = 6 cm. Positive v indicates image is behind the mirror (virtual). Magnification m = -v/u = -6/(-10) = 0.6 (positive, so erect). Image is virtual, erect, and diminished, located 6 cm behind the mirror.
Q13. The magnification produced by a plane mirror is +1. What does this mean?
Answer:
Magnification +1 means the image formed by a plane mirror is of the same size as the object (size ratio 1:1) and the positive sign indicates that the image is erect (upright) and virtual.
Q14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a concave mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer:
Given: Object height h = 5.0 cm, object distance u = -20 cm, radius of curvature R = 30 cm for concave mirror. Focal length f = -R/2 = -15 cm (negative for concave mirror). Mirror formula: 1/v + 1/u = 1/f => 1/v + 1/(-20) = 1/(-15) => 1/v = -1/15 + 1/20 = (-4+3)/60 = -1/60 => v = -60 cm. Negative v indicates image is in front of the mirror (real). Magnification m = -v/u = -(-60)/(-20) = 60/(-20) = -3. Image height h' = m × h = -3 × 5 = -15 cm. Negative height indicates inverted image. Image is real, inverted, and enlarged (size 15 cm), located 60 cm in front of the mirror.
Q15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed to get a sharp image? Find the nature and size of the image.
Answer:
Given: Object height h = 7.0 cm, object distance u = -27 cm, focal length f = -18 cm (concave mirror). Mirror formula: 1/v + 1/u = 1/f => 1/v + 1/(-27) = 1/(-18) => 1/v = -1/18 + 1/27 = (-3+2)/54 = -1/54 => v = -54 cm. Screen should be placed 54 cm in front of the mirror. Magnification m = -v/u = -(-54)/(-27) = 54/(-27) = -2. Image height h' = m × h = -2 × 7 = -14 cm. Image is real, inverted, and enlarged (size 14 cm).
Q16. Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer:
Given power P = -2.0 D (dioptre). Focal length f (in metres) = 1/P = 1/(-2.0) = -0.5 m = -50 cm. The negative focal length indicates it is a concave lens (diverging lens).
Q17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Answer:
Given power P = +1.5 D. Focal length f = 1/P = 1/1.5 = 10/15 = 2/3 ? 0.67 m = 66.7 cm. Positive power indicates a converging lens (convex lens).