A plano-convex lens is made of refractive index
1.6. The radius of curvature of the curved surface
is 60 cm. The focal length of the lens is??
Answers
Answer:
it's 30 cm
pls mark me as brainliest
Answer:
8CM
➝ This question says that a plane convex lens of focal length 16 cm is to be made of glass of refraction. Index 1.5 We have to find the radius of curvature of the curved surface.
Given that
➝ Focal length of plane convex lens is 16 cm
➝ Index is 1.5
To find
➝ The radius of curvature of the curved surface.
Solution
➝ The radius of curvature of the curved surface = 8 cm
Usingconcept
➝ Len's maker formula
Using formula
{\bold{\sf{\longmapsto Len's \: maker = \dfrac{1}{f} = (_{m}u_{I}-1) [\dfrac{1}{R_1} - \dfrac{1}{R_2}}}}⟼Len
′
smaker=
f
1
=(
m
u
I
−1)[
R
1
1
−
R
2
1
{\large{\bold{\sf{\underline{These \; denotes}}}}}
Thesedenotes
➝ {\sf_{m}u_{l}}
m
u
l
means refractive index of lens, respect to the medium.
➝ {\sf{\dfrac{1}{R_1}}}
R
1
1
means radius of curvature of the first surface of lens.
➝ {\sf{\dfrac{1}{R_2}}}
R
2
1
means radius of curvature of the second surface of lens.
➝ f means focal length
{\large{\bold{\sf{\underline{Solution}}}}}
Solution
~ Here we have –
➝ {\sf_{m}u_{l}}
m
u
l
as 1.5
➝ {\sf{\dfrac{1}{R_1}}}
R
1
1
as ?
➝ {\sf{\dfrac{1}{R_2}}}
R
2
1
as ∞
➝ Focal length of lens as 16 cm
~ Finding {\sf{\dfrac{1}{R_1}}}
R
1
1
{\bold{\sf{\longmapsto Len's \: maker = \dfrac{1}{f} = (_{m}u_{I}-1) [\dfrac{1}{R_1} - \dfrac{1}{R_2}}}}⟼Len
′
smaker=
f
1
=(
m
u
I
−1)[
R
1
1
−
R
2
1
{\bold{\sf{\longmapsto Len's \: maker = \dfrac{1}{16} = (1.5 - 1) [\dfrac{1}{R_1} - \dfrac{1}{\infin}]}}}⟼Len
′
smaker=
16
1
=(1.5−1)[
R
1
1
−
∞
1
]
{\bold{\sf{\longmapsto Len's \: maker = \dfrac{1}{16} = (0.5) [\dfrac{1}{R_1} - 0]}}}⟼Len
′
smaker=
16
1
=(0.5)[
R
1
1
−0]
{\bold{\sf{\longmapsto Len's \: maker = \dfrac{1}{16} = (0.5) \times \dfrac{1}{R_1}}}}⟼Len
′
smaker=
16
1
=(0.5)×
R
1
1
{\bold{\sf{\longmapsto R_{1} = 0.5 \times 16}}}⟼R
1
=0.5×16
{\bold{\sf{\longmapsto R_{1} = 8 \: cm}}}⟼R
1
=8cm