NCERT Exercise - Q10
Textbook Exercise Problem
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Find the position, size and the nature of the image formed.
Step-by-Step Verified Solution:
- Step 1: Sign convention: Converging lens (convex) focal length f = +10 cm. Object distance u = -25 cm. Object height h = +5 cm.
- Step 2: Lens formula: 1/f = 1/v - 1/u => 1/v = 1/f + 1/u.
- Step 3: 1/v = 1/10 + 1/(-25) = (5 - 2) / 50 = 3/50 => v = +50/3 = +16.67 cm.
- Step 4: Magnification m = v / u = (+50/3) / (-25) = -2/3.
- Step 5: Image height h' = m × h = (-2/3) × 5 = -10/3 ≈ -3.33 cm.
- Step 6: Nature: Positive v indicates image is formed on the opposite side; negative h' indicates REAL and INVERTED.
Final Answer: Position: v = +16.67 cm, Size: 3.33 cm, Nature: Real and Inverted
Key Concept / Exam Strategy: In Cartesian sign convention, object distance u is always negative.
Generated Exercise #1
Foundation
Question: Find the focal length of a lens of power -2.0 D. What type of lens is this?
Core Concept Tested: Power to focal length inverse relation and sign interpretation
Mentor Hint: f = 1 / P. Negative power means concave diverging lens.
Step-by-Step Solution:
- Step 1: P = -2.0 D. Formula: f = 1 / P meters.
- Step 2: f = 1 / (-2.0) = -0.5 m = -50 cm.
- Step 3: Since focal length is negative, it is a Concave (diverging) lens used for correcting myopia.
Verified Answer: Focal length = -0.5 m (-50 cm), Type: Concave lens
Generated Exercise #2
Intermediate
Question: Light enters from air to a glass plate having refractive index 1.50. What is the speed of light in the glass? (Speed of light in vacuum c = 3 × 10⁸ m/s).
Core Concept Tested: Absolute refractive index velocity definition
Mentor Hint: n = c / v => v = c / n.
Step-by-Step Solution:
- Step 1: Formula: n = c / v.
- Step 2: v = c / n = (3 × 10⁸) / 1.50 = 2 × 10⁸ m/s.
Verified Answer: 2 × 10⁸ m/s
Generated Exercise #3
Intermediate
Question: A convex mirror used for rearview on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position and nature of the image.
Core Concept Tested: Convex mirror virtual diminished image formation
Mentor Hint: R = +3 m => f = +1.5 m. u = -5 m. 1/f = 1/v + 1/u.
Step-by-Step Solution:
- Step 1: f = R / 2 = +1.50 m. u = -5.00 m.
- Step 2: 1/v = 1/f - 1/u = 1/1.5 - (-1/5) = 1/1.5 + 1/5 = 10/15 + 3/15 = 13/15.
- Step 3: v = +15/13 = +1.15 m behind the mirror.
- Step 4: Magnification m = -v/u = - (1.15) / (-5.00) = +0.23 (Virtual, erect, diminished).
Verified Answer: v = +1.15 m behind mirror; Virtual, erect, diminished (m = +0.23)
Generated Exercise #4
JEE-NEET Foundation
Question: Two thin lenses of powers +3.5 D and -1.5 D are placed in close contact. Find the net power and focal length of this combination.
Core Concept Tested: Algebraic superposition of thin lens optical power
Mentor Hint: P_net = P1 + P2; f_net = 1 / P_net.
Step-by-Step Solution:
- Step 1: P_net = P1 + P2 = +3.5 D + (-1.5 D) = +2.0 D.
- Step 2: f_net = 1 / P_net = 1 / 2.0 = +0.5 m = +50 cm.
- Step 3: The combination behaves as a converging (convex) lens of focal length 50 cm.
Verified Answer: Net Power = +2.0 D, Focal Length = +50 cm (Converging)
Generated Exercise #5
Advanced HOTS
Question: At what distance from a concave mirror of focal length 20 cm should an object be placed so that an image magnified 2 times is formed: (a) Real, (b) Virtual?
Core Concept Tested: Dual branch magnification analysis in concave spherical optics
Mentor Hint: For real image, m = -2; for virtual image, m = +2.
Step-by-Step Solution:
- Step 1: Given f = -20 cm.
- Step 2: Case (a) Real image: m = -v/u = -2 => v = 2u. Using 1/f = 1/v + 1/u: -1/20 = 1/(2u) + 1/u = 3/(2u) => 2u = -60 => u = -30 cm.
- Step 3: Case (b) Virtual image: m = -v/u = +2 => v = -2u. Using 1/f = 1/v + 1/u: -1/20 = -1/(2u) + 1/u = 1/(2u) => 2u = -20 => u = -10 cm.
Verified Answer: (a) Real image: u = 30 cm in front of mirror; (b) Virtual image: u = 10 cm in front of mirror