The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.
Solution:
The given quantities are
Height of the object h = 5 cm
Object distance u = -25 cm
Focal length f = 10 cm
The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.
The image distance is the distance between the position of convex lens and the position where the image is formed.
We know that the ‘focal length’ of a convex lens can be found using the below formula
1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
f
1
=
v
1
−
u
1
Here f is the focal length, v is the image distance which is known to us and u is the object distance.
The image height can be derived from the magnification equation, we know that
Magnification=h′h=vu\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=
h
h
′
=
u
v
Thus,
h′h=vu\frac{h^{\prime}}{h}=\frac{v}{u}
h
h
′
=
u
v
First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,
1f=1v−1(−25)\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}
f
1
=
v
1
−
(−25)
1
1v=1f+1(−25)=110−125\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}
v
1
=
f
1
+
(−25)
1
=
10
1
−
25
1
1v=25−10250=15250\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}
v
1
=
250
25−10
=
250
15
v=25015=503=16.66 cmv=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}v=
15
250
=
3
50
=16.66 cm
Then using the magnification relation, we can get the image height as follows
h′5=−16.6625\frac{h^{\prime}}{5}=-\frac{16.66}{25}
5
h
′
=−
25
16.66
So, the image height will be
h′=−5×16.6625=−3.332 cmh^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}h
′
=−5×
25
16.66
=−3.332 cm
Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.
