Question

If the two thin convex lenses each of focal length 8 cm are kept coaxially and separated by a distance 20 cm each other, then equivalent focal
length is
Options​

Answer 1

Answer:

Answer-

( identity : (a-b)² = a²-2ab +b²)

= {(2x - 5y)}^{2}=(2x−5y)

2

= {(2x)}^{2} - 2(2x)(5y) + {(5y)}^{2}=(2x)

2

−2(2x)(5y)+(5y)

2

= 4 {x}^{2} - 20xy + 25 {y}^{2}=4x

2

−20xy+25y

2

Answer 2

Answer:

            Equivalent focal length D =-20CM      

Explanation:

TO FIND:

      Equivalent focal length is  

    GIVEN:

          The two thin convex lenses each of focal length 8 cm are kept coaxially .

     Separated by a distance 20 cm each other.

                    $F= 8CM$

                     $u=-12$

             Lens formula $\frac{1}{f}= \frac{1}{v} -\frac{1}{u}$

                                 $\frac{1}{v}= \frac{1}{f}+ \frac{1}{u} \\ =\frac{1}{8} -\frac{1}{12}\\ =0.04166$

                          v=0.04166

                  Equivalent focal length = $\frac{1}{f_{1} }+ \frac{1}{f_{2} }- \frac{d}{f_{1} f_{2} }$

                                                          $\frac{1}{F} =\frac{1}{12}+ \frac{1}{12} -\frac{20}{12*12}$

                                                         $D=\frac{d_{f} }{f_{1} }$

                                                           $D=\frac{(20)(-12)}{12}$

                                                              $D= -20CM$

                            Equivalent focal length D =-20CM        

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