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JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments

उत्तल-दर्पण की फोकस दूरी निकालने के एक प्रयोग में निम्न डाटा प्राप्त हुआ
बिंब उत्तल लैंस उत्तल दर्पण प्रतिबिंब
  \(22.2\,cm\)   \(32.2\,cm\)   \(45.8\,cm\)   \(71.2\,cm\)
उत्तल लैंस की फोकस दूरी \(f_{1}\) तथा उत्तल-दर्पण की फोकस दूरी \(f_{2}\) है। index correction नगण्य है। तब

  1. A \(f_1 = 7.8\,cm\,\,\,\,\, f_2 = 12.7\,\,cm\)
  2. B \(f_1 = 12.7\,cm\,\,\,\,\, f_2 = 7.8\,\,cm\)
  3. C \(f_1 = 15.6\,cm\,\,\,\,\, f_2 = 25.4\,\,cm\)
  4. D \(f_1 = 7.8\,cm\,\,\,\,\, f_2 = 25.4\,\,cm\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(f_1 = 7.8\,cm\,\,\,\,\, f_2 = 12.7\,\,cm\)

Step-by-step Solution

Detailed explanation

From lens \(u_{1}=-(32.2-22.2) c m=-10 c m\) \(v_{1}=(71.2-32.2) \mathrm{cm}=39 \mathrm{cm}\) \(\frac{1}{f_{1}}=\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{36}+\frac{1}{10}=\frac{49}{390}\) or \(f_{1}=\frac{390}{49} \mathrm{cm}=7.8 \mathrm{cm}\) For mirror…
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