JEE MAIN

<div><div><li> <p>The aperture diameter of a telescope is 5 m. The separation between the moon and the earth is 4 × 10<sup>5</sup> Km. With light of wavelength of 5500 Å, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to</p> </li></div></div>

<div><p>Given: The aperture diameter of a telescope is 
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      5 
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      m 
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     a = 5\text{\ }m 
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 </math></p><p>wavelength of incident light is 
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    <mi>
      λ 
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      5500 
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      Å 
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   <annotation encoding="application/x-tex">
     \lambda = 5500\ \text{Å} 
   </annotation> 
  </semantics> 
 </math>,</p><p>distance between moon and earth is 
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      4 
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     D = 4 \times 
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 </math> 
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       10 
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       5 
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     10^{5}\text{\ }km 
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  </semantics> 
 </math>.</p><p>To find: 
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     d 
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 </math>, the minimum separation between two objects on the surface of moon, so they can be resolved using this telescope.</p><p> <img src="http://test-master-question-data.s3.ap-south-1.amazonaws.com/questionImg/912c4a43-43d4-4aba-a433-51d911e638b1.jpg" style="width:2.27236in;height:1.15864in" alt=""> </p><p>Minimum angle for clear resolution:</p><p> 
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      1.22 
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     \theta = 1.22\frac{\lambda}{a} 
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  </semantics> 
 </math> </p><p> 
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     d = \theta D 
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 </math> </p><p> 
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      1.22 
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     d = 1.22\frac{\lambda}{a}D 
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  </semantics> 
 </math> </p><p> 
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        1.22 
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        5500 
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        × 
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         10 
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          − 
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          10 
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        4 
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     = \frac{1.22 \times 5500 \times 10^{- 10} \times 4 \times 10^{5} \times 10^{3}}{5} 
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 </math> </p><p> 
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      1.22 
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      11 
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      × 
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      4 
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      = 
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      53.68 
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      ≈ 
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    <mn>
      60 
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     = 1.22 \times 11 \times 4 = 53.68 \approx 60\text{\ }m 
   </annotation> 
  </semantics> 
 </math> </p></div>