The focal length of concave lens is 15cm at what distance object should be kept so that image formed will be 10cm away from lens. Find magnification of object.

AcademicPhysicsNCERTClass 10

Given:

Focal length of concave lens, $f$ = $-$15 cm   $(focal\ length\ of\ a\ concave\ lens\ is\ always\ taken\ negative)$


‚ÄčImage distance, $v$ = $-$10 cm                         $(image\ distance\ is\ taken\ negative\ because\ image\ formed\ by\ a\ concave\ lens\ is\ on\ the\ left\ side\ of\ the\ lens)$


To find: Object distance $(u)$, and magnification $(m)$.


Solution:

From the lens formula, we know that-


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


Substituting the given values in the formula, we get-


$\frac {1}{(-10)}-\frac {1}{u}=\frac {1}{(-15)}$


$-\frac {1}{10}-\frac {1}{u}=-\frac {1}{15}$


$\frac {1}{u}=\frac {1}{15}-\frac {1}{10}$


$\frac {1}{u}=\frac {2-3}{30}$


$\frac {1}{u}=\frac {-1}{30}$


$u=-\frac {1}{30}$


$u=-30cm$


Thus, the object is at a distance of 30 cm from the concave lens, and the negative sign implies that it is on the left side of it. 


Now,


Foo magnification we know that-


$m=\frac {v}{u}$


Substituting the given values in the formula, we get-


$m=\frac {-10}{-30}$


$m=\frac {1}{3}$


$m=0.33$


Thus, the magnification $m$, produced by the lens is 0.33 which is less than 1, therefore the image will be smaller than the object.

raja
Updated on 10-Oct-2022 13:25:47

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