Q1: A thumb pin is positioned at a distance of 15 cm from a convex mirror of a focal length of 20 cm. Determine the position and nature of the image.
Given:
Object distance, u = -15 cm (sign convention: object distance is negative)
Focal length, f = +20 cm (convex mirror has positive focal length)
Formula: Mirror formula: 1/f = 1/v + 1/u
1/v = 1/f - 1/u
Calculation:
1/v = 1/20 - 1/(-15)
1/v = 1/20 + 1/15
1/v = (3 + 4)/60
1/v = 7/60
v = 60/7 = 8.57 cm
Nature of image:
Positive value of v indicates image is formed behind the mirror.
Therefore, image is virtual, erect, and diminished.
Answer: Image position = 8.57 cm behind the mirror, Virtual and erect
Object distance, u = -15 cm (sign convention: object distance is negative)
Focal length, f = +20 cm (convex mirror has positive focal length)
Formula: Mirror formula: 1/f = 1/v + 1/u
1/v = 1/f - 1/u
Calculation:
1/v = 1/20 - 1/(-15)
1/v = 1/20 + 1/15
1/v = (3 + 4)/60
1/v = 7/60
v = 60/7 = 8.57 cm
Nature of image:
Positive value of v indicates image is formed behind the mirror.
Therefore, image is virtual, erect, and diminished.
Answer: Image position = 8.57 cm behind the mirror, Virtual and erect
Q2: An image of a specimen appears to be 11.5 cm behind a concave mirror with a focal length of 13.5 cm. Find the specimen's distance from the mirror.
Given:
Image distance, v = +11.5 cm (behind the mirror, so virtual image, positive for concave mirror)
Focal length, f = -13.5 cm (concave mirror has negative focal length)
Formula: Mirror formula: 1/f = 1/v + 1/u
1/u = 1/f - 1/v
Calculation:
1/u = 1/(-13.5) - 1/11.5
1/u = -1/13.5 - 1/11.5
1/u = -(11.5 + 13.5)/(13.5 × 11.5)
1/u = -25/(155.25)
1/u = -0.161
u = -6.21 cm
Answer: Object distance = 6.21 cm in front of the mirror
Image distance, v = +11.5 cm (behind the mirror, so virtual image, positive for concave mirror)
Focal length, f = -13.5 cm (concave mirror has negative focal length)
Formula: Mirror formula: 1/f = 1/v + 1/u
1/u = 1/f - 1/v
Calculation:
1/u = 1/(-13.5) - 1/11.5
1/u = -1/13.5 - 1/11.5
1/u = -(11.5 + 13.5)/(13.5 × 11.5)
1/u = -25/(155.25)
1/u = -0.161
u = -6.21 cm
Answer: Object distance = 6.21 cm in front of the mirror
Q3: A convex mirror used for rear-view on an automobile has a radius of curvature of 4.00 m. If a bus is located at 5.00 m from this mirror, find the image's position, nature, and size.
Given:
Radius of curvature, R = +4.00 m
Focal length, f = R/2 = +2.00 m (convex mirror)
Object distance, u = -5.00 m
Formula: Mirror formula: 1/f = 1/v + 1/u
1/v = 1/f - 1/u
Calculation for image position:
1/v = 1/2 - 1/(-5)
1/v = 1/2 + 1/5
1/v = (5 + 2)/10
1/v = 7/10
v = 10/7 = 1.428 m
Nature of image:
Positive v indicates image is formed behind the mirror (virtual, erect)
Magnification: m = -v/u = -1.428/(-5) = +0.2856
Image is diminished (0.2856 times the object size)
Answer: Image position = 1.428 m behind the mirror, Virtual, erect, and diminished
Radius of curvature, R = +4.00 m
Focal length, f = R/2 = +2.00 m (convex mirror)
Object distance, u = -5.00 m
Formula: Mirror formula: 1/f = 1/v + 1/u
1/v = 1/f - 1/u
Calculation for image position:
1/v = 1/2 - 1/(-5)
1/v = 1/2 + 1/5
1/v = (5 + 2)/10
1/v = 7/10
v = 10/7 = 1.428 m
Nature of image:
Positive v indicates image is formed behind the mirror (virtual, erect)
Magnification: m = -v/u = -1.428/(-5) = +0.2856
Image is diminished (0.2856 times the object size)
Answer: Image position = 1.428 m behind the mirror, Virtual, erect, and diminished
Q4: An object is placed 15 cm away from a converging lens of a focal length of 10 cm. Determine the position, size, and nature of the image formed.
Given:
Object distance, u = -15 cm (sign convention: u is negative)
Focal length, f = +10 cm (converging/convex lens)
Formula: Lens formula: 1/f = 1/v - 1/u
1/v = 1/f + 1/u
Calculation for image position:
1/v = 1/10 + 1/(-15)
1/v = 1/10 - 1/15
1/v = (3 - 2)/30
1/v = 1/30
v = 30 cm
Magnification: m = v/u = 30/(-15) = -2
Negative sign indicates inverted image.
|m| = 2 means image is twice the size of object.
Nature: Real, inverted, and magnified (since v is positive and m is negative)
Answer: Image position = 30 cm on the opposite side, Real, inverted, and magnified (2 times)
Object distance, u = -15 cm (sign convention: u is negative)
Focal length, f = +10 cm (converging/convex lens)
Formula: Lens formula: 1/f = 1/v - 1/u
1/v = 1/f + 1/u
Calculation for image position:
1/v = 1/10 + 1/(-15)
1/v = 1/10 - 1/15
1/v = (3 - 2)/30
1/v = 1/30
v = 30 cm
Magnification: m = v/u = 30/(-15) = -2
Negative sign indicates inverted image.
|m| = 2 means image is twice the size of object.
Nature: Real, inverted, and magnified (since v is positive and m is negative)
Answer: Image position = 30 cm on the opposite side, Real, inverted, and magnified (2 times)
Q5: A concave lens of focal length 20 cm forms an image 15 cm from the lens. Determine the power of a lens. Also, how far is the object positioned from the lens?
Given:
Focal length, f = -20 cm = -0.20 m (concave lens has negative focal length)
Image distance, v = -15 cm = -0.15 m (for concave lens, image is virtual and on same side)
Formula for Power: P = 1/f (in meters)
Calculation of Power:
P = 1/(-0.20) = -5 D
Formula for object distance: Lens formula: 1/f = 1/v - 1/u
1/u = 1/v - 1/f
Calculation for object distance:
1/u = 1/(-15) - 1/(-20)
1/u = -1/15 + 1/20
1/u = (-4 + 3)/60
1/u = -1/60
u = -60 cm
Answer: Power = -5 D, Object distance = 60 cm from the lens
Focal length, f = -20 cm = -0.20 m (concave lens has negative focal length)
Image distance, v = -15 cm = -0.15 m (for concave lens, image is virtual and on same side)
Formula for Power: P = 1/f (in meters)
Calculation of Power:
P = 1/(-0.20) = -5 D
Formula for object distance: Lens formula: 1/f = 1/v - 1/u
1/u = 1/v - 1/f
Calculation for object distance:
1/u = 1/(-15) - 1/(-20)
1/u = -1/15 + 1/20
1/u = (-4 + 3)/60
1/u = -1/60
u = -60 cm
Answer: Power = -5 D, Object distance = 60 cm from the lens
Q6: The angle of incidence for a ray of light from air to water interface is 40°. If the ray travels through the water with a refractive index of 1.33, calculate the angle of refraction.
Given:
Angle of incidence, i = 40°
Refractive index of water, n = 1.33
Refractive index of air, nₐ = 1
Formula: Snell's Law: nₐ sin i = n sin r
sin r = (nₐ sin i) / n
Calculation:
sin r = (1 × sin 40°) / 1.33
sin 40° = 0.6428
sin r = 0.6428 / 1.33
sin r = 0.4833
r = sin⁻¹(0.4833)
r = 28.8°
Answer: Angle of refraction = 28.8°
Angle of incidence, i = 40°
Refractive index of water, n = 1.33
Refractive index of air, nₐ = 1
Formula: Snell's Law: nₐ sin i = n sin r
sin r = (nₐ sin i) / n
Calculation:
sin r = (1 × sin 40°) / 1.33
sin 40° = 0.6428
sin r = 0.6428 / 1.33
sin r = 0.4833
r = sin⁻¹(0.4833)
r = 28.8°
Answer: Angle of refraction = 28.8°
