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  • An object 1point 5 cm in size is placed on the side of the convex lens in the arrangement a above The distance between the object and the convex lens is 40cm Determine the magnification produced by the two-lens system and the size of the image

Q 9.21 (b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40cm. Determine the magnification produced by the two-lens system, and the size of the image.

Answers (1)

best_answer

Given: 

Object height = 1.5 cm 

Object distance from convex lens = -40cm

According to the lens formula:

\frac{1}{v_{fromconvex}}=\frac{1}{f_{convex}}+\frac{1}{u_{fromconvex}}

\frac{1}{v_{fromconvex}}=\frac{1}{30}+\frac{1}{-40}=\frac{1}{120}

v_{fromconvex}=120

Magnification due to convex lens:

 m_{convex}=-\frac{v}{u}=-\frac{120}{-40}=3

The image of convex lens will act as an object for concave lens,

so,

\frac{1}{v_{fromconcave}}=\frac{1}{f_{concave}}+\frac{1}{u_{fromconcave}}

u_{concave}= 120 - 8 = 112

\frac{1}{v_{fromconcave}}= \frac{1}{-20}+\frac{1}{112}

\frac{1}{v_{fromconcave}}= -\frac{92}{2240}

v_{fromconcave}= \frac{-2240}{92}

Magnification due to concave lens :

m_{concave}= \frac{2240}{92}*\frac{1}{112}=\frac{20}{92}

The combined magnification is:

 m_{combined}=m_{convex}*m_{concave}

m_{combined}=3*\frac{20}{92}=0.652

Hence height of the image = m_{combined}*h 

= 0.652 * 1.5 = 0.98cm

Hence, the height of the image is 0.98cm.

Posted by

Pankaj Sanodiya

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