Answer: D!
It is the option with the greatest amplitude.
Answer:
a) m =1 θ = sin⁻¹ λ / d, m = 2 θ = sin⁻¹ ( λ / 2d)
, c) m = 3
Explanation:
a) In the interference phenomenon the maxima are given by the expression
d sin θ = m λ
the maximum for m = 1 is at the angle
θ = sin⁻¹ λ / d
the second maximum m = 2
θ = sin⁻¹ ( λ / 2d)
the third maximum m = 3
θ = sin⁻¹ ( λ / 3d)
the fourth maximum m = 4
θ = sin⁻¹ ( λ / 4d)
b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.
I = I₀ cos² (Ф) (sin x / x)²
Ф = π d sin θ /λ
x = pi a sin θ /λ
where a is the width of the slits
with the values of part a are introduced in the expression and we can calculate intensity of each maximum
c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present
maximum interference d sin θ = m λ
first diffraction minimum a sin θ = λ
we divide the two expressions
d / a = m
In our case
3a / a = m
m = 3
order three is no longer visible
Answer:
Is that an actual question for school?
Answer:
first lens v = 48 cm
second lens v = -15.6 cm
magnification = 1.67
final image is virtual
and final image is upright
Explanation:
given data
distance = 16 cm
focal length f1 = 12 cm
focal length f2 = 10.0 cm
to find out
location of the final image and magnification and Type of image
solution
we apply here lens formula that is
1/f = 1/v + 1/u .....................1
put here all value and find v for 1st lens
1/12 = 1/v + 1/16
v = 48 cm
and find v for 2nd lens
here u = 20- 48 = -28
- 1/10 = 1/v - 1/28
v = -15.6 cm
and
magnification = first lens (v/u) × second lens ( v/u)
magnification = (-15.6/-28) × ( 48/16)
magnification = 1.67
so here final image is virtual
and final image is upright
Answer:
Explanation:
Magnetic field due to a long solenoid at the center is given by
here we know that
N = 300
L = 30 cm
i = 12 A
now magnetic field is given as
Now magnetic flux through the disc is given as
