(a) Using the ray diagram for a system of two lenses of focal lengthsf_{1} and f_{2} in contact with each other, show that the two lens system can be regarded as equivalent to a single lens of focal length f, where \frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}} . Also write the relation for the equivalent power of the lens combination.

(b) Determine the position of the image formed by the lens combination given in the figure.

    

 

 
 
 

Answers (1)
S safeer

(a)   Let us consider the two thin lenses A and B forming the image I and I_{1} respectively, from the object (O) For I the distance is u  and for I_{1} the distance is v_{1}

        For lens 'A', the lens formula is ;

            \frac{1}{f_{1}}=\frac{1}{v}-\frac{1}{u} _______(1)

    Where v is the image distance

                 u is the object distance

    For lens 'B' , the lens formula is :    

        \frac{1}{f_{2}}=\frac{1}{v}-\frac{1}{v_{1}} ________(2)       

On adding both equations, we get 
       \frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{v_{1}}-\frac{1}{u}+\frac{1}{v}-\frac{1}{v_{1}}

    \frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{v}-\frac{1}{u}            \left [ here, \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \right ]

then, 

        \frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}

Therefore, the equivalent power form here would be 

        P=P_{1}+P_{2}

(b)  The image formed by the lense of f=+10cm

        \frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}

        \frac{1}{v_{1}}-\frac{1}{30}=\frac{1}{10}

     \therefore v_{1}=15cm

This image formed by the lens act as an object from the concave lens.

        Therefore,

        u_{2}=15-5=10cm

       \frac{1}{f_{2}}+\frac{1}{v_{2}}=\frac{1}{u_{2}}

        \frac{1}{-10}=\frac{1}{v}-\frac{1}{10}

here, v=\infty

   therefore, virtual image forms at the right of the concave lens at v=\infty and act as an object for convex lens (f=+30cm)

        \frac{1}{v_{3}}-\frac{1}{\infty}=\frac{1}{f_{3}}

        v_{3}=30cm

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