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A : For interference fringes to be seen,...

A : For interference fringes to be seen, the condition `(s)/(S) lt (lambda)/(d)` should be satisfied. Where .s. be the size of source slit, S is its distance from plane of two slits and .d. is the distance between two slits.
R : In Y.D.S.E, if distance of source slit from the two slits (s) decreases, the interference pattern gets more sharp.

A

Both A and R are true and R is the correct explanation of A

B

Both A and R are true and R is not the correct explanation of A

C

A is true and R is false

D

Both A and R are false

Text Solution

AI Generated Solution

The correct Answer is:
To solve the given problem, we need to analyze the two statements provided: Assertion (A) and Reason (R). ### Step 1: Analyze the Assertion (A) The assertion states that for interference fringes to be seen, the condition \( \frac{s}{S} < \frac{\lambda}{d} \) must be satisfied, where: - \( s \) is the size of the source slit, - \( S \) is the distance from the source slit to the plane of the two slits, - \( d \) is the distance between the two slits, - \( \lambda \) is the wavelength of the light used. This condition implies that the size of the source slit relative to its distance from the slits must be small compared to the wavelength divided by the distance between the slits. If this condition is not satisfied, the interference pattern will not be clearly observable. **Conclusion for Assertion:** The assertion is **True**. ### Step 2: Analyze the Reason (R) The reason states that in Young's Double Slit Experiment (Y.D.S.E), if the distance of the source slit from the two slits (S) decreases, the interference pattern gets sharper. To analyze this, we note that if \( S \) decreases, the ratio \( \frac{s}{S} \) increases (since \( s \) remains constant). If \( \frac{s}{S} \) increases, it may violate the condition \( \frac{s}{S} < \frac{\lambda}{d} \). This means that the interference pattern may become less distinct or even disappear if the condition is not satisfied. **Conclusion for Reason:** The reason is **False** because decreasing \( S \) does not lead to a sharper interference pattern; instead, it can lead to a failure to observe the interference pattern altogether. ### Final Conclusion - Assertion (A) is **True**. - Reason (R) is **False**. Thus, the correct option is that the assertion is true, but the reason is false.
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AAKASH SERIES-WAVES OPTICS-EXERCISE -IB (ASSERTION AND REASON)
  1. Assertion The pattern and position of fringes always remain same even ...

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  2. A : Y.D.S.E, as the source slit width increases, fringe pattern gets l...

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  3. A : For interference fringes to be seen, the condition (s)/(S) lt (lam...

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  4. Statement-1: In Young's double slit experiment the two slits are at di...

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  5. A : When the coherent sources are far apart, interference pattern cann...

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  6. Assertion : When a thin transparent sheet is placed in front of both t...

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  7. A : The soap film in sun light is colourful. R : Thin films produce ...

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  8. A : Thin films such as soap bubble or a thin layer of oil on water sho...

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  9. Assertion:- Newton's rings are formed in the reflected system. When th...

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  10. Assertion (A) : The film which appears bright in reflected system will...

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  11. A : If a thin soap film is arranged vertically the spectrum of coloure...

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  12. Radio waves diffract around building, although light waves do not. The...

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  13. A : It is impossible to see an object as small as an atom regardless o...

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  14. A : Diffraction is common is sound but not common in light waves R :...

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  15. A : There is no specific important physical difference between interfe...

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  16. A : At the first glance, the top surf Morpho butterfly.s wing appears ...

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  17. A : Coloured spectrum is seen when we look through a muslin cloth. R...

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  18. Assertion : The clouds in sky generally appear to be whitish. Reason...

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  19. A : In double slit experiment, the pattern on the screen is actually a...

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  20. Assertion : Standard optical diffraction gratings cannot be used for ...

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