refractive index of lens is 1.5 when an object is placed at 30cm image is formed at 20cm.find it focal length which lens is it ? if the radio of curvature are equal that what is it's value?
Answers
Answer:
Given,
Refractive index of the lens, μ
g
=1.5
Radius of curvature of first plano-concave lens, R
1
=20 cm
Radius of curvature of second plano-concave lens, R
2
=30 cm
Refractive index of the liquid μ
l
= 34
Using Lens makers formula,
f
1=(μ−1)( R11 - R21)
Now, the system can be considered as 3 lenses in contact as shown in the figure.
Focal length of lens 1
f 11 =( 23−1)( ∞1 − 201 )=( 21 )( 20-1 )= 40−1
f 1 =−40 cm
Focal length of lens 2f 21=( 34 −1)( 201−301 )=( 31 )( 605 )= 1805
f 2=36 cm
Focal length of lens 3f 31 =( 23−1)( −301−1 )=( 21 )( 30−1 )= 60−1
f 3 =−60 cm
The equivalent focal length of the combination is given as
F
1
=
f
1
1
+
f
2
1
+
f
3
1
F
1
=
40
−1
+
36
1
+
60
−1
F
1
=
360
−9+10−6
=
360
−5
F=
−5
360
=−72 cm
∴ The system will behave as a Concave Lens of focal length 72 cm.
Answer:
Given,
Refractive index of the lens, μ
g
=1.5
Radius of curvature of first plano-concave lens, R
1
=20 cm
Radius of curvature of second plano-concave lens, R
2
=30 cm
Refractive index of the liquid μ
l
=
3
4
Using Lens makers formula,
f
1
=(μ−1)(
R
1
1
−
R
2
1
)
Now, the system can be considered as 3 lenses in contact as shown in the figure.
Focal length of lens 1
f
1
1
=(
2
3
−1)(
∞
1
−
20
1
)=(
2
1
)(
20
−1
)=
40
−1
f
1
=−40 cm
Focal length of lens 2
f
2
1
=(
3
4
−1)(
20
1
−
−30
1
)=(
3
1
)(
60
5
)=
180
5
f
2
=36 cm
Focal length of lens 3
f
3
1
=(
2
3
−1)(
−30
1
−
∞
1
)=(
2
1
)(
30
−1
)=
60
−1
f
3
=−60 cm
The equivalent focal length of the combination is given as
F
1
=
f
1
1
+
f
2
1
+
f
3
1
F
1
=
40
−1
+
36
1
+
60
−1
F
1
=
360
−9+10−6
=
360
−5
F=
−5
360
=−72 cm
∴ The system will behave as a Concave Lens of focal length 72 cm.
Hence, the correct answer is OPTION B.