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GATE Electronics and Communications (EC) 2019 Solved Paper
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© examsnet.com
Question : 40 of 65
Marks:
+1
,
-0
A germanium sample of dimensions
1
cm
×
1
cm
1\ \text{cm} \times 1\ \text{cm}
1
cm
×
1
cm
is illuminated with a
20
mW
,
600
nm
20\ \text{mW}, 600\ \text{nm}
20
mW
,
600
nm
laser light source as shown in the figure. The illuminated sample surface has a 100 nm of loss-less Silicon dioxide layer that reflects one-fourth of the incident light. From the remaining light, one-third of the power is reflected from the Silicon dixodieGermanium interface, one-third is absorbed in the Germanium layer, and one-third is transmitted through the other side of the sample. If the absorption coefficient of Germanium at
600
nm
600\ \text{nm}
600
nm
is
3
×
3 \times
3
×
1
0
4
cm
−
1
10^{4}\ \text{cm}^{-1}
1
0
4
cm
−
1
and the bandgap is
0.66
eV
0.66\ \text{eV}
0.66
eV
, the thickness of the Germanium layer, rounded off to 3 decimal places, is ____________
μ
m
\mu\ \text{m}
μ
m
.
Your Answer:
Validate
Solution:
1
−
e
−
α
x
=
0.5
1-e^{-\alpha x}=0.5
1
−
e
−
αx
=
0.5
e
−
α
x
=
0.5
e^{-\alpha x}=0.5
e
−
αx
=
0.5
now
∝
=
3
×
1
0
4
cm
−
1
\propto = 3 \times 10^{4}\ \text{cm}^{-1}
∝=
3
×
1
0
4
cm
−
1
∴
x
=
−
ln
(
0.5
)
3
×
1
0
4
\therefore x = \frac{-\ln(0.5)}{3 \times 10^{4}}
∴
x
=
3
×
1
0
4
−
l
n
(
0.5
)
© examsnet.com
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