SATHEE: Optics 6 Question 5

Optics 6 Question 5

6. In a compound microscope, the intermediate image is

$(2000, 2 M)$

(a) virtual, erect and magnified

(b) real, erect and magnified

(d) virtual, erect and reduced

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Answer:

Correct Answer: 6. (a)

Solution:

Here, wavelength of light used $(λ) = 5303 \AA$ .

Distance between two slit $(d) = 19.44 μ - m$

Width of single slit $(a) = 4.05 μ m$

Here, angular width between first and second diffraction minima

$θ ≃ \frac{λ}{a}$

and angular width of a fringe due to double slit is

$θ^{'} = \frac{λ}{d}$

$∴$ Number of fringes between first and second diffraction minima, $n = \frac{θ}{θ^{'}} = \frac{\frac{λ}{a}}{\frac{λ}{d}} = \frac{d}{a} = \frac{19.44}{4.05} = 4.81$ or $n ≃ 5$

$∴ 5$ interfering bright fringes lie between first and second diffraction minima.

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Optics 6 Question 5

6. In a compound microscope, the intermediate image is

Answer:

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