Answer to Question #173257 in Optics for zainab zahra

Question

Consider a con\"1ging lens of focal length /. When an object is placed 60 cm away from the !en.., a real image is produced. \\'hen the object is mm'lld to 40 cm away from the lens, the image III0\"'5 10 cm further nway from the lens tb!la it wns wbea the object was at 60 cm. \Vhnt is /

Answer 1

ORIGINAL QUESTION

when an object is placed 60.0 cm from a certain convrging lens, it forms a real image. When the object is movd to 40.0 cm from the lens, the image moves 10.0 cm farthar from the lens. Find the focal (f) length of this lens. (SOURCE: GOOGLE AND EDUCATIONAL WEBSITE)

ANSWER:

LET,

Initial position of object$=u_1=-60cm$

final position of object$\\=u_2=-40cm$

initial position of image$\\=v_1=v$

final position of image$=v_2=v+10$

ACCORDING TO LENS FORMULA:

$\boxed{{1\over v}-{1\over u}={1\over f}}$

putting values for 1^st condition;

${1\over v_1}-{1\over u_1}={1\over f}$

$\\={1\over v}-{1\over -60}={1\over f}$ ...................(1)

putting values for 1^st condition;

${1\over v_2}-{1\over u_2}={1\over f}$

$\\={1\over v+1}-{1\over -40}={1\over f}$ ......................(2)

now equating equation(1) and (2)...

$\implies{1\over v}-{1\over -60}={1\over v+1}-{1\over -40}$

$\implies{1\over v}-{1\over v+1}={1\over 40}-{1\over 60}$

$\implies v^2+10v=1200$

v=30 or -40 (-40 not considered because image position is +ve side)

$\therefore\boxed{ v=30}$

NOW ; PUTTING TIS VALUE IN EQUATION (1)....we get....

$\implies{1\over 30}-{1\over -60}={1\over f}$

$\implies {1\over f}={1\over 20}$

$\boxed{\therefore f= 20 cm}$