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The two lenses of a compound microscope ...

The two lenses of a compound microscope are of focal lengths 2 cm and 5 cm. If an object is placed at a distance of 2.1 cm from the objetive of focal length 2 cm the final image forms at the least distance of distinct vision of a normal eye. Find the frnagnifying powerjof the microscope

A

20

B

6

C

120

D

60

Text Solution

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The correct Answer is:
To find the magnifying power of the compound microscope with given focal lengths and object distance, we can follow these steps: ### Step 1: Identify the given values - Focal length of the objective lens (f₀) = 2 cm - Focal length of the eyepiece lens (fₑ) = 5 cm - Object distance from the objective lens (u₀) = -2.1 cm (negative as per sign convention) ### Step 2: Use the lens formula for the objective lens The lens formula is given by: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \] Where: - \( v₀ \) = image distance from the objective lens - \( u₀ \) = object distance from the objective lens - \( f₀ \) = focal length of the objective lens Substituting the values into the lens formula: \[ \frac{1}{v₀} - \frac{1}{-2.1} = \frac{1}{2} \] This simplifies to: \[ \frac{1}{v₀} + \frac{1}{2.1} = \frac{1}{2} \] ### Step 3: Solve for \( v₀ \) Rearranging gives: \[ \frac{1}{v₀} = \frac{1}{2} - \frac{1}{2.1} \] Finding a common denominator (which is 42): \[ \frac{1}{v₀} = \frac{21 - 20}{42} = \frac{1}{42} \] Thus, \[ v₀ = 42 \text{ cm} \] ### Step 4: Calculate the magnifying power of the objective lens (M₀) The magnifying power of the objective lens is given by: \[ M₀ = \frac{v₀}{u₀} \] Substituting the values: \[ M₀ = \frac{42}{2.1} = 20 \] ### Step 5: Calculate the total magnification (M) The total magnification of the compound microscope is given by: \[ M = M₀ \left(1 + \frac{D}{fₑ}\right) \] Where: - \( D \) = least distance of distinct vision (approximately 25 cm for a normal eye) - \( fₑ \) = focal length of the eyepiece lens = 5 cm Substituting the values: \[ M = 20 \left(1 + \frac{25}{5}\right) = 20 \left(1 + 5\right) = 20 \times 6 = 120 \] ### Final Answer The magnifying power of the microscope is **120**. ---

To find the magnifying power of the compound microscope with given focal lengths and object distance, we can follow these steps: ### Step 1: Identify the given values - Focal length of the objective lens (f₀) = 2 cm - Focal length of the eyepiece lens (fₑ) = 5 cm - Object distance from the objective lens (u₀) = -2.1 cm (negative as per sign convention) ### Step 2: Use the lens formula for the objective lens ...
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