An object 60 cm tall is placed on the principal axis of a convex lens. Its 20 cm tall image is formed on the screen placed at a distance of 10 cm from the lens. Calculate the focal length of the lens.
Answers
Answer:
The ‘focal length’ of convex length is 7.14 cm.
Explanation:
The given quantities are
Height of the object h = 50 cm
Height of the image formed h’ = -20 cm
Image distance v = 10 cm
The image distance is the distance between the position of convex lens and the position where the image is formed.
The ‘focal length’ of a convex lens can be found using the below formula
\bold{\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 object distance is the distance between the ‘object position’ and the lens position. To determine the focal length, first we should find the object distance.
From the magnification equation, we know that
\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=
h
h
′
=
u
v
Thus,
\begin{gathered}\begin{aligned} \frac{h^{\prime}}{h} &=\frac{v}{u} \\ \\ \frac{-20}{50} &=\frac{10}{u} \end{aligned}\end{gathered}
h
h
′
50
−20
=
u
v
=
u
10
So, the object distance will be
u=-10 \times \frac{50}{20}=-25 \mathrm{cm}u=−10×
20
50
=−25cm
So the focal length will be
\frac{1}{f}=\frac{1}{10}-\frac{1}{(-25)}=\frac{1}{10}+\frac{1}{25}
f
1
=
10
1
−
(−25)
1
=
10
1
+
25
1
\frac{1}{f}=\frac{25+10}{250}=\frac{35}{250}
f
1
=
250
25+10
=
250
35
f=\frac{250}{35}=\frac{50}{7}=7.14 \mathrm{cm}f=
35
250
=
7
50
=7.14cm
Thus the focal length of the convex lens is 7.14 cm
Answer:
- 5 Cm
Explanation:
magnification factor
M= height of image/ height of object
M= 20/60
M=1/3
NOW M=F/(F-D०)
1/3=F/F-10
F-10=3F
2F= - 10
f=-5 cm