JEE MAIN - Physics (2017 (Offline) - No. 2)
A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15cm from a converging
lens of magnitude of focal length 20cm. A beam of parallel light falls on the diverging lens. The final
image formed is:
real and at a distance of 6 cm from the convergent lens.
real and at a distance of 40 cm from convergent lens.
virtual and at a distance of 40 cm from convergent lens.
real and at a distance of 40 cm from the divergent lens.
Giải thích
As parallel beam incident on diverging lens so the image will be formed
at the focus of diverging lens.
$$ \therefore $$ v = –25 cm_en_2_1.png)
The image formed by diverging lens is used as an object for converging lens.
So for converging lens
u = –25 – 15 = –40 cm,
f = 20 cm
So Final image formed by converging lens
$${1 \over {20}} = {1 \over V} - {1 \over { - 40}}$$
$$ \Rightarrow $$ V = 40 cm
V is positive so image will be real and will form at right side of converging lens at 40 cm.
$$ \therefore $$ v = –25 cm
The image formed by diverging lens is used as an object for converging lens.
So for converging lens
u = –25 – 15 = –40 cm,
f = 20 cm
So Final image formed by converging lens
$${1 \over {20}} = {1 \over V} - {1 \over { - 40}}$$
$$ \Rightarrow $$ V = 40 cm
V is positive so image will be real and will form at right side of converging lens at 40 cm.