યંગના ડબલ સ્લીટના પ્રયોગમાં બે સ્લિટ વચ્ચેનું અંતર સમય સાથે \({d}({t})={d}_{0}+{a}_{0}\, sin\omega \,t\) મુજબ બદલાય છે; જ્યાં \({d}_{0}, \omega\) અને \(a_{0}\) અચળાંકો છે. સ્થિર છે. સમય સાથે મેળવેલી સૌથી મોટી શલાકાની પહોળાઈ અને સૌથી નાની શલાકાની પહોળાઈ વચ્ચેનો તફાવત કઈ રીતે આપવામાં આવે?

A \(\frac{\lambda D}{d_{0}+a_{0}}\)
B \(\frac{2 \lambda \cdot D a_{0}}{\left(d_{0}^{2}-a_{0}^{2}\right)}\)
C \(\frac{2 \lambda {D}\left({d}_{0}\right)}{\left({d}_{0}^{2}-{a}_{0}^{2}\right)}\)
D \(\frac{\lambda D}{d_{0}^{2}} a_{0}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{2 \lambda \cdot D a_{0}}{\left(d_{0}^{2}-a_{0}^{2}\right)}\)

Step-by-step Solution

Detailed explanation

Fringe Width, \(\beta=\frac{\lambda D }{ d }\) \(\beta_{\max } \Rightarrow d _{\min } \text { and } \beta_{\min } \Rightarrow d _{\max }\) \(d = d _{0}+ a _{0} \sin \omega t\) \(d _{\max }= d _{0}+ a _{0} \text { and } d _{\min }= d _{0}- a _{0}\)…

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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