વસ્તુ અને પડદા વચ્ચેનું અંતર \(100\,cm\) છે. લેન્સના બે સ્થાન માટે વસ્તુનું વાસ્તવિક પ્રતિબિંબ પડદા પર પડે છે. બે સ્થાન વ્ચ્ચેનું અંતર \(40\,cm\) છે. લેન્સનો પાવર \(\left(\frac{ N }{100}\right) D\) હોય તો \(N\) ........

A \(445\)
B \(495\)
C \(486\)
D \(476\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(476\)

Step-by-step Solution

Detailed explanation

Using displacement method \(f =\frac{ D ^{2}- d ^{2}}{4 D }\) Here, \(D=100 cm\) \(d =40 cm\) \(f =\frac{100^{2}-40^{2}}{4(100)}=21 cm\) \(P =\frac{1}{ f }=\frac{100}{21} D \quad \frac{ N }{100}=\frac{100}{21} N =476\)

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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