Chapter 9: Light - Reflection and Refraction
Key Formulas:
• Mirror Formula: 1/v + 1/u = 1/f
• Magnification: m = -v/u = h'/h
• Lens Formula: 1/v - 1/u = 1/f
• Power of lens: P = 1/f (in metres) = 100/f (in cm)
• Snell's Law: n₁ sin i = n₂ sin r
Q1. Define the principal focus of a concave mirror.
Solution
The principal focus of a concave mirror is a point on its principal axis where all the light rays, initially 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?
Solution
Given: R = 20 cm
f = R/2 = 20/2 = 10 cm
The focal length is 10 cm.
Q3. Name a mirror that can give an erect and enlarged image of an object.
Solution
A concave mirror can give an erect and enlarged image when the object is placed between the pole and the focus of the mirror.
Q4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Solution
We prefer a convex mirror because:
• It gives an erect (upright) image
• It gives a diminished (smaller) image
• It provides a wider field of view, allowing the driver to see more traffic behind
Q5. A concave mirror produces three times magnified real image of an object placed at 10 cm in front of it. Where is the image located?
Solution
Given: u = -10 cm, m = -3 (real image)
m = -v/u
-3 = -v/(-10)
v = -30 cm
The image is located at 30 cm in front of the mirror.
Q6. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Solution
Given: R = +32 cm (convex mirror)
f = R/2 = 32/2 = +16 cm
The focal length is +16 cm.
Q7. A concave mirror produces an image of size double that of the object placed at a distance of 15 cm from it. Find the focal length of the mirror.
Solution
Given: u = -15 cm, m = -2 (real image, inverted, double size)
m = -v/u
-2 = -v/(-15)
v = -30 cm
1/v + 1/u = 1/f
1/(-30) + 1/(-15) = 1/f
-1/30 - 1/15 = 1/f
-1/30 - 2/30 = 1/f
-3/30 = 1/f
f = -10 cm
The focal length is -10 cm (concave mirror).
Q8. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position of the image.
Solution
Given: u = -10 cm, f = +15 cm
1/v + 1/u = 1/f
1/v + 1/(-10) = 1/15
1/v - 1/10 = 1/15
1/v = 1/15 + 1/10
1/v = 2/30 + 3/30
1/v = 5/30
v = +6 cm
The image is located at 6 cm behind the mirror (virtual image).
Q9. A convex lens has a focal length of 10 cm. At what distance from the lens should the object be placed so as to obtain a real and inverted image of double the size?
Solution
Given: f = +10 cm, m = -2
m = v/u → v = -2u
1/v - 1/u = 1/f
1/(-2u) - 1/u = 1/10
-1/(2u) - 1/u = 1/10
-1/(2u) - 2/(2u) = 1/10
-3/(2u) = 1/10
u = -15 cm
The object should be placed at 15 cm from the lens.
Q10. What is the power of a concave lens of focal length 2 m?
Solution
Given: f = -2 m (concave lens)
P = 1/f = 1/(-2) = -0.5 D
The power is -0.5 Dioptre.
Q11. A doctor prescribes a corrective lens of power +1.5 D. Find the focal length of the lens. Is it a converging or diverging lens?
Solution
Given: P = +1.5 D
f = 1/P = 1/1.5 = 0.67 m = 67 cm
The focal length is 67 cm. Since the power is positive, it is a converging (convex) lens.
Q12. 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 so as to get a sharp image? Find the size of the image.
Solution
Given: h = 7 cm, u = -27 cm, f = -18 cm
1/v + 1/u = 1/f
1/v + 1/(-27) = 1/(-18)
1/v = -1/18 + 1/27
1/v = (-3 + 2)/54
1/v = -1/54
v = -54 cm
m = -v/u = -(-54)/(-27) = -2
h' = m × h = -2 × 7 = -14 cm
The screen should be placed at 54 cm from the mirror. The image size is 14 cm (inverted).