JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
The distance between an object and a screen is \(100\, cm .\) A lens can produce real image of the object on the screen for two different positions between the screen and the object. The distance between these two positions is \(40\, cm .\) If the power of the lens is close to \(\left(\frac{ N }{100}\right) D\) where \(N\) is an integer, the value of \(N\) is.......
- A \(445\)
- B \(495\)
- C \(486\)
- D \(476\)
Answer & Solution
Correct Answer
(D) \(476\)
Step-by-step Solution
Detailed explanation
Using displacement method \(f =\frac{ D ^{2}- d ^{2}}{4 D }\) Here, \(D=100 cm\) \(d =40 cm\) \(f =\frac{100^{2}-40^{2}}{4(100)}=21 cm\) \(P =\frac{1}{ f }=\frac{100}{21} D \quad \frac{ N }{100}=\frac{100}{21} N =476\)
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