Question

A distant hot air balloon subtends 0.25" at the objective lens of a telescope. The Image subtends 1.5 at eyepiece when viewed with relaxed eye. The objective and eyepiece are in a 35 cm long tube Find the focal length of eyepiece lens (in cm).​

Answer 1

Answer:

Here, f0=100cm,fe=2cm

α=0.5∘,β=?

As m=fofe=βα

∴β=fofeα=1002×0.5∘=25∘

Explanation:

Answer 2

Answer:

The focal length of eyepiece is 5.83cm

Explanation:

Given the angle subtended at the objective, $\theta_{0} =0.25^{o}$

                                      and at the image, $\theta_{i} =1.5^{o}$

Angular magnification of telescope is given by

                                      $m=\frac{\theta_{i} }{\theta_{0} }= \frac{f_{o} }{f_{e} }$

                                    $\implies \frac{1.5 }{0,25 }= \frac{f_{o} }{f_{e} }\implies\frac{f_{o} }{f_{e} }=5$

                                                            $\implies f_{o}=5{f_{e}$              ....(1)

And also the length of the tube, $d=35 cm$

where length of the tube is given by  $d=f_{o} +f_{e}$

                                                      $\implies35 =f_{o} +f_{e}$        

Using equation (1),                      

                               $35 =5f_{e} +f_{e}\implies35 =6f_{e}$

Hence the focal length of eyepiece is,

                                     $f_{e} =\frac{35}{6} =5.83cm$