Wave Interference

Step 1: Concept
Fringe width .
Step 2: Analysis
New distance and .
New width .
Step 3: Conclusion
The fringe width becomes 4 times its original value.
Final Answer: (C)

Step 1: Moles of Reactant
Millimoles of .
Step 2: Stoichiometry
From , 25 millimoles of produce 25 millimoles of .
Step 3: Milliequivalents
is bivalent ( ), so milliequivalents .
Step 4: Acid Volume
Volume of 0.1 N HCl .
Final Answer: (c)

Step 1: Understanding the interference pattern.
In an interference experiment, the fringe width ( ) is given by the formula: where: - is the wavelength of light, - is the distance between the slits and the screen, - is the distance between the two slits.

Step 2: Substituting known values.
From the question: - , - , - .
Using the formula for fringe width:

Step 3: Conclusion.
The fringe width is . Therefore, the correct answer is (2).

Step 1: Understanding the Concept:
Einstein's photoelectric equation: .
Step 2: Detailed Explanation:
, .

=
= .
Step 3: Final Answer:
Thus, .

Step 1: Understanding the Concept:
Path difference at point on the circle: . For maxima: .
Step 2: Detailed Explanation:
. Condition: . So, . : → 2 points. : same 2 points. For : each gives 4 points → total = . : gives 2 points. Total points: .
Step 3: Final Answer:
Number of maxima on the circle .

Step 1: Understanding the Concept:
For thin film in air, condition for constructive interference: . Since film is on glass ( ), both reflections undergo a phase change of .
Step 2: Detailed Explanation:
Condition: . Substitute: . Solve: . Find thickness:
Step 3: Final Answer:

Concept: Fringe width in YDSE:

Step 1: Fringe width depends directly on wavelength:

Step 2: Blue light has smaller wavelength than orange light:

Step 3: Therefore, the distance between consecutive fringes will decrease.