Answer:
The lenses with different focal length are four.
Explanation:
Given that,
Radius of curvature R₁= 4
Radius of curvature R₂ = 8
We know ,
Refractive index of glass = 1.6
When, R₁= 4, R₂ = 8
We need to calculate the focal length of the lens
Using formula of focal length
Put the value into the formula
When , R₁= -4, R₂ = 8
Put the value into the formula
When , R₁= 4, R₂ = -8
Put the value into the formula
When , R₁= -4, R₂ = -8
Put the value into the formula
Hence, The lenses with different focal length are four.
First
let us imagine the projectile launched at initial velocity V and at angle
θ relative to the horizontal. (ignore wind resistance)
Vertical component y:
The
initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum
height of h, the vertical velocity
will be 0, therefore the time t taken to attain this maximum height is:
h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g
where
g is acceleration due to gravity
Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike
the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the
horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g]
D = (2V²sinθ cosθ)/g
D = (V²sin2θ)/g
In order for D (horizontal distance) to be
maximum, dD/dθ = 0
That is,
2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2 or θ = π/4
Therefore it is now<span> shown that the maximum horizontal travelled is attained when
the launch angle is π/4 radians, or 45°.</span>
Acceleration is any change in speed or direction of motion.
The dimension of speed is [length/time],
so a change is [length/time²].
Popular units include [meter/second²] and [feet/second²] .
________________________
Direction almost always boils down to an angle, (which technically
has no dimensions), so a change in direction is [angle/time] .
Popular units include [radian/second] and [degree/second] .
This is an interesting (read tricky!) variation of Rydberg Eqn calculation.
Rydberg Eqn: 1/λ = R [1/n1^2 - 1/n2^2]
Where λ is the wavelength of the light; 1282.17 nm = 1282.17×10^-9 m
R is the Rydberg constant: R = 1.09737×10^7 m-1
n2 = 5 (emission)
Hence 1/(1282.17 ×10^-9) = 1.09737× 10^7 [1/n1^2 – 1/25^2]
Some rearranging and collecting up terms:
1 = (1282.17 ×10^-9) (1.09737× 10^7)[1/n2 -1/25]
1= 14.07[1/n^2 – 1/25]
1 =14.07/n^2 – (14.07/25)
14.07n^2 = 1 + 0.5628
n = √(14.07/1.5628) = 3
