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An isotropic point source of light is kept at distance of 60 cm from a screen, the intensity of light obtained on center is I1. Now a lens of focal length 20 cm is introduced at exactly midway between source and screen and intensity obtained is I2. Find I2/8I1.
Correct answer is '2'. Can you explain this answer?
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An isotropic point source of light is kept at distance of 60 cm from a...
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An isotropic point source of light is kept at distance of 60 cm from a...
Given:
- Distance between the isotropic point source of light and the screen = 60 cm
- Intensity of light obtained on the center of the screen without the lens = I1
- Focal length of the lens = 20 cm
- Intensity of light obtained on the center of the screen with the lens = I2
To Find:
I2/8I1
Solution:
1. Without the Lens:
When there is no lens, the light from the isotropic point source will spread out uniformly in all directions. Therefore, the intensity of light obtained on the screen will decrease with increasing distance.
2. With the Lens:
When the lens is introduced midway between the source and the screen, it acts as a converging lens. The lens will converge the light rays and focus them on the screen. This will result in a higher intensity of light on the screen compared to the case without the lens.
3. Calculation of I2:
To calculate I2, we can use the lens formula:
1/f = 1/v - 1/u
Where,
f = focal length of the lens
v = image distance from the lens
u = object distance from the lens
In this case, the object distance u is equal to half the distance between the source and the screen, i.e., 60 cm / 2 = 30 cm.
Substituting the values into the lens formula:
1/20 = 1/v - 1/30
Solving this equation, we get v = 60 cm.
Therefore, the image distance from the lens is 60 cm.
Now, the intensity of light on the screen can be calculated using the magnification formula:
m = v/u
Where,
m = magnification
v = image distance from the lens
u = object distance from the lens
Substituting the values into the magnification formula:
m = 60 cm / 30 cm = 2
Therefore, the magnification of the lens is 2.
Since intensity is directly proportional to the square of the magnification, we can calculate I2 as:
I2 = (m^2) * I1
= (2^2) * I1
= 4I1
4. Calculation of I2/8I1:
To calculate I2/8I1, we substitute the values of I2 and I1:
I2/8I1 = (4I1) / (8I1)
= 4/8
= 1/2
= 0.5
Therefore, I2/8I1 = 0.5.
Conclusion:
The intensity ratio I2/8I1 is equal to 0.5 or 1/2.
- Distance between the isotropic point source of light and the screen = 60 cm
- Intensity of light obtained on the center of the screen without the lens = I1
- Focal length of the lens = 20 cm
- Intensity of light obtained on the center of the screen with the lens = I2
To Find:
I2/8I1
Solution:
1. Without the Lens:
When there is no lens, the light from the isotropic point source will spread out uniformly in all directions. Therefore, the intensity of light obtained on the screen will decrease with increasing distance.
2. With the Lens:
When the lens is introduced midway between the source and the screen, it acts as a converging lens. The lens will converge the light rays and focus them on the screen. This will result in a higher intensity of light on the screen compared to the case without the lens.
3. Calculation of I2:
To calculate I2, we can use the lens formula:
1/f = 1/v - 1/u
Where,
f = focal length of the lens
v = image distance from the lens
u = object distance from the lens
In this case, the object distance u is equal to half the distance between the source and the screen, i.e., 60 cm / 2 = 30 cm.
Substituting the values into the lens formula:
1/20 = 1/v - 1/30
Solving this equation, we get v = 60 cm.
Therefore, the image distance from the lens is 60 cm.
Now, the intensity of light on the screen can be calculated using the magnification formula:
m = v/u
Where,
m = magnification
v = image distance from the lens
u = object distance from the lens
Substituting the values into the magnification formula:
m = 60 cm / 30 cm = 2
Therefore, the magnification of the lens is 2.
Since intensity is directly proportional to the square of the magnification, we can calculate I2 as:
I2 = (m^2) * I1
= (2^2) * I1
= 4I1
4. Calculation of I2/8I1:
To calculate I2/8I1, we substitute the values of I2 and I1:
I2/8I1 = (4I1) / (8I1)
= 4/8
= 1/2
= 0.5
Therefore, I2/8I1 = 0.5.
Conclusion:
The intensity ratio I2/8I1 is equal to 0.5 or 1/2.
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An isotropic point source of light is kept at distance of 60 cm from a screen, the intensity of light obtained on center is I1. Now a lens of focal length 20 cm is introduced at exactly midway between source and screen and intensity obtained is I2. Find I2/8I1.Correct answer is '2'. Can you explain this answer?
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An isotropic point source of light is kept at distance of 60 cm from a screen, the intensity of light obtained on center is I1. Now a lens of focal length 20 cm is introduced at exactly midway between source and screen and intensity obtained is I2. Find I2/8I1.Correct answer is '2'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An isotropic point source of light is kept at distance of 60 cm from a screen, the intensity of light obtained on center is I1. Now a lens of focal length 20 cm is introduced at exactly midway between source and screen and intensity obtained is I2. Find I2/8I1.Correct answer is '2'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An isotropic point source of light is kept at distance of 60 cm from a screen, the intensity of light obtained on center is I1. Now a lens of focal length 20 cm is introduced at exactly midway between source and screen and intensity obtained is I2. Find I2/8I1.Correct answer is '2'. Can you explain this answer?.
An isotropic point source of light is kept at distance of 60 cm from a screen, the intensity of light obtained on center is I1. Now a lens of focal length 20 cm is introduced at exactly midway between source and screen and intensity obtained is I2. Find I2/8I1.Correct answer is '2'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An isotropic point source of light is kept at distance of 60 cm from a screen, the intensity of light obtained on center is I1. Now a lens of focal length 20 cm is introduced at exactly midway between source and screen and intensity obtained is I2. Find I2/8I1.Correct answer is '2'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An isotropic point source of light is kept at distance of 60 cm from a screen, the intensity of light obtained on center is I1. Now a lens of focal length 20 cm is introduced at exactly midway between source and screen and intensity obtained is I2. Find I2/8I1.Correct answer is '2'. Can you explain this answer?.
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