યંગના વ્યતિકરણના પ્રયોગમાં વપરાયેલ પ્રકાશની તરંગલંબાઈ 500 nm છે. સ્લિટથી પડદાનું અંતર 100 cm અને બે સ્લિટ વચ્ચેનું અંતર 1 mm છે, તો પાંચમી અને ત્રીજી પ્રકાશિત શલાકા વચ્ચેનું અંતર કેટલું હશે ?

A 3 mn
B 2 mm
C 1 mm
D 4 mm

Verified Solution

Answer & Solution

Correct Answer

(C) 1 mm

Step-by-step Solution

Detailed explanation

C.1 mm
\(\lambda=500 mm\)
\(D =100 cm\)
\(d =1 mm\)
શલાકા વચ્ચેનું અંતર
\(=\frac{5 \lambda D }{d}-\frac{3 \lambda D }{d}\)
\(=\frac{2 \lambda D }{d}\)
\(=\frac{2 \times 5 \times 10^{-7} \times 1}{1 \times 10^{-3}}\)
\(=10^{-3} m\)
\(=1 mm\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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