SATHEE: Optics Question 164

Optics Question 164

Question: A circular disc is placed in front of a narrow source. When the point of observation is at a distance of 1 meter from the disc, then the disc covers first HPZ. The intensity at this point is I0. The intensity at a point distance 25 cm from the disc will be

Options:

A) $I_{1} = 0.531 I_{0}$

B) $I_{1} = 0.053 I_{0}$

C) $I_{1} = 53 I_{0}$

D) $I_{1} = 5.03 I_{0}$

Show Answer

Answer:

Correct Answer: A

Solution:

$I_{0} = R^{2} = \frac{R_{2}^{2}}{4}$ .

Number of HPZ covered by the disc at $b = 25 c m$

$n_{1} b_{1} = n_{2} b_{2}$

$n_{2} = \frac{n_{1} b_{1}}{b_{2}} + \frac{1 \times 1}{0.25} = 4$ .

Hence the intensity at this point is

$I = {R^{'}}^{2} = {(\frac{R_{5}}{2})}^{2} = {(\frac{R_{5}}{R_{4}} \times \frac{R_{4}}{R_{3}} \times \frac{R_{3}}{R_{2}})}^{2} \times {(\frac{R_{2}}{2})}^{2}$ or $1 = {[0.9]}^{6}$

$I_{1} = 0.531 I_{0}$ Hence the correct answer will be (a).

Optics Question 164

Question: A circular disc is placed in front of a narrow source. When the point of observation is at a distance of 1 meter from the disc, then the disc covers first HPZ. The intensity at this point is I0. The intensity at a point distance 25 cm from the disc will be

Options:

Answer:

Solution:

Engineering Preparation

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Optics Question 164

Question: A circular disc is placed in front of a narrow source. When the point of observation is at a distance of 1 meter from the disc, then the disc covers first HPZ. The intensity at this point is I0. The intensity at a point distance 25 cm from the disc will be

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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