A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
1. Given Data
Using the Cartesian sign convention for a concave mirror:
- Object Size (h1): +2.5 cm
- Object Distance (u): -27 cm
- Radius of Curvature (R): -36 cm
2. Calculations
Step A: Find Focal Length (f)
f = R / 2 = -36 / 2 = -18 cm
Step B: Find Image Position (v)
Using the mirror formula (1/v + 1/u = 1/f):
1/v = 1/f - 1/u
1/v = 1/(-18) - 1/(-27)
1/v = -1/18 + 1/27
1/v = (-3 + 2) / 54
1/v = -1 / 54
v = -54 cm
Step C: Find Size and Nature
Using the magnification formula (m = -v/u = h2/h1):
m = -(-54) / (-27) = -2
h2 = m × h1
h2 = -2 × 2.5 = -5 cm
Final Answer
- Screen Position: 54 cm in front of the mirror.
- Nature: Real and Inverted (since magnification is negative).
- Image Size: 5 cm (Magnified).
3. What happens if the candle moves closer?
Currently, the object is placed between the Center (C) and Focus (F).
Case A: Moving closer to Focus
If the candle is moved closer to the mirror (but stays beyond 18 cm), the image will move away from the mirror and get larger. The screen would have to be moved further back.
Case B: Moving closer than Focus (< 18 cm)
If the candle is moved closer than 18 cm (between the Pole and Focus), the image becomes Virtual and Erect. A virtual image forms behind the mirror and cannot be caught on a screen.
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