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Geometrical Optics — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsGeometrical Optics1 Mark
Question
A screen is placed a distance 40cm away from an illuminated object. A converging lens is placed between the source and the screen and it is attempted to form the image of the source on the screen. If no position could be found, the focal length of the lens:
  1. Must be less than 10cm.
  2. Must be greater than 20cm.
  3. Must not be greater than 20cm.
  4. Must not be less than 10cm.
✓

Answer

  1. Must be greater than 20cm.
Explanation:
$\text{v}=(40-4)$
$\frac{1}{\text{f}}=\frac{1}{40-4}-\frac{1}{(-\text{u})}$
$\frac{\text{df}}{\text{du}}=0$ for f minimum.
$\frac{\text{df}}{\text{du}}=1-\frac{\text{u}}{20}=0$
$\text{u}=20$
$\text{f}_{\text{min}}=10\text{cm}$

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