# The magnitude of (|E| / |H|) in a uniform plane wave is:

1. $$\sqrt {\mu \epsilon}$$
2. Infinity
3. $$\sqrt {\dfrac{\mu}{ \epsilon}}$$
4. 1

## Answer (Detailed Solution Below)

Option 3 : $$\sqrt {\dfrac{\mu}{ \epsilon}}$$

## Parallel Plane Waveguide MCQ Question 1 Detailed Solution

Concept:

• Electric field $$\left( {\vec E} \right)$$ & magnetic field $$\left( {\vec H} \right)$$ are both orthogonal/transverse to each other as well as for the direction of propagation, called transverse electromagnetic wave (TEM).
• E & H combination called uniform plane wave because E & H has some magnitude through any transverse plane.

Intrinsic Impedance $$\left( \eta \right) = \frac{E}{H}$$

And also $$\eta = \sqrt {\frac{{j\omega \mu }}{{\sigma + j\omega \varepsilon }}}$$

$$\Rightarrow \eta = \frac{E}{H} = \sqrt {\frac{\mu }{\varepsilon }}$$

Pointing vector is a vector whose direction is the direction of wave propagation pointing vector $$= \vec E \times \vec H$$ (Hence it is a direction of wave propagation)

Where,

E = Electric filed

B = magnetic field

H = Magnetic field

μ0 = Permeability of free space = 4π x 10-7 H / m

ϵ0 = Permittivity of free space = 8.85 x 10-12 F/m

# The dispersion equation of a waveguide, which relates the wavenumber k to the frequency ω, is $$k\left( \omega \right) = \left( {1/\;c} \right)\;\sqrt {{\omega ^2} - \omega _o^2}$$ where the speed of light c = 3 × 108 m/s, and ωo is a constant. If the group velocity is 2 × 108 m/s, then the phase velocity is

1. 1.5 × 108 m/s
2. 2 × 108 m/s
3. 3 × 108 m/s
4. 4.5 × 108 m/s

## Answer (Detailed Solution Below)

Option 4 : 4.5 × 108 m/s

## Parallel Plane Waveguide MCQ Question 2 Detailed Solution

Concept:

Phase velocity:

The rate at which the phase of the wave propagates in space

$${V_p} = \frac{\omega }{\beta }$$

Group Velocity:

The velocity with which overall envelope of the wave travels

$${V_g} = \frac{{d\omega }}{{d\beta }} = \frac{{d\omega }}{{dk}}$$

$$K\left( \omega \right) = \frac{1}{c}\sqrt {\omega - \omega _0^2} \ldots \left( 1 \right)$$

Vg = 2 × 108 m/s (green)

$${V_g} = \frac{{d\omega }}{{dk}}$$

$$\frac{{dk}}{{d\omega }} = \frac{1}{{{V_g}}}$$

differentiating (1)

$$\frac{{dk}}{{d\omega }} = \frac{1}{{{V_g}}} = \frac{{1 \times 2\omega }}{{2C\sqrt {{\omega ^2} - \omega _0^2} }} = \frac{1}{{{V_g}}}$$

$$\Rightarrow \frac{\omega }{{3 \times {{10}^8}\sqrt {{\omega ^2} - \omega _0^2} }} = \frac{1}{{2 \times {{10}^8}}}$$

$$\sqrt {{\omega ^2} - \omega _0^2} = \frac{{2\omega }}{3}$$

$${V_p} = \frac{\omega }{K} = \frac{{\omega c}}{{\sqrt {{\omega ^2} - \omega _0^2} }}$$

$$= \frac{{\omega c}}{{\left( {\frac{{2\omega }}{3}} \right)}}$$

$${V_p} = \frac{{3c}}{2} = \frac{{3 \times {{10}^8} \times 3}}{2} = 4.5 \times {10^8}m/s$$

Shortcut:

Group velocity × Phase velocity = c2

2 × 108 × phase velocity = (3 × 108)2

$${V_p} = \frac{{9 \times {{10}^{16}}}}{{2 \times {{10}^8}}}$$

= 4.5 × 108 m/s

# The velocity of electromagnetic waves in a dielectric medium with relative permittivity of 4 will

1. 3 × 108 m/s
2. 6 × 108 m/s
3. 1.5 × 108 m/s
4. 0.75 × 108 m/s

## Answer (Detailed Solution Below)

Option 3 : 1.5 × 108 m/s

## Parallel Plane Waveguide MCQ Question 3 Detailed Solution

Concept:

The velocity (Vp) of a plane electromagnetic wave is given as-

$${V_p} = \frac{1}{{\sqrt {μ \varepsilon } }}$$

Here, μ = μoμr

ε = εo εr

$${v_p} = \frac{1}{{\sqrt {{μ _o}{μ _r}{\varepsilon _o}{\varepsilon _r}} }}$$

$${v_p} = \frac{c}{{\sqrt {{μ _r}{\varepsilon _r}} }}$$-----(1)

$$c = \frac{1}{{\sqrt {{μ _o}{\varepsilon _o}} }} = 3 \times {10^8}\;m/sec$$

μ0 = permeability in free space 4π × 10-7 H/m

εo = permittivity in free space 8.854 × 10-12 C2/Nm2

Calculation:

Given:

μr = 4

From Equation (1):

$$v_p=\frac{3\times10^8}{\sqrt4}$$

vp = 1.5 x 108 m/sec

# For a parallel plate transmission line, let v be the speed of propagation and Z be the characteristic impedance. Neglecting fringe effects, a reduction of spacing between the plates by a factor of two results in

1. halving of v and no change in Z
2. no change in v and halving of Z
3. no change in both v and Z
4. halving of both v and Z

## Answer (Detailed Solution Below)

Option 2 : no change in v and halving of Z

## Parallel Plane Waveguide MCQ Question 4 Detailed Solution

Concept:

Consider the following parallel plate line:

Assuming w ≫ h, the fringing effect can be ignored.

Where:

The bold line represents the direction of the Electric Field, and the dotted lines represent the direction of the Magnetic Field.

V = Ey h

I = Hx w

The Characteristic Impedance is given by:

$$Z = \frac{V}{I} = \frac{{{E_y}}}{{{H_x}}} \cdot \frac{h}{w} = \eta \frac{h}{w}$$

$$Z = \eta \frac{h}{w}$$        ---(1),

η = intrinsic impedance of dielectric medium.

If ‘h’ is reduced by a factor of 2 then Z will be halved.

Alternate Method:

$$Z = \sqrt {\frac{L}{C}}$$

$$Z = \sqrt {\frac{{\frac{{\mu h}}{w}}}{{\frac{{Ew}}{h}}}}$$

$$Z = \sqrt {\frac{\mu }{E}} \;\frac{h}{w}$$

We observe that If ‘h’ is halved, z is halved.

Important Point:

If the student has learned the formula for L and C then this problem will be easily solved by:

$$Z = \sqrt {\frac{L}{C}} \;\;and\;\;v = \frac{1}{{\sqrt {LC} }}$$

Also, the formula for L and C for the following are also important:

Coaxial cable:

$$C = \frac{{2\pi El}}{{In\;\left( {\frac{b}{a}} \right)}}$$

$$h = \frac{{\mu l}}{{2\pi }}In\;\left( {\frac{b}{a}} \right)$$

Parallel wires:

$$C = \frac{{\pi El}}{{In\left( {\frac{D}{r}} \right)}}$$

r = radius of the wire

$$L = \frac{{ul}}{\pi }In\left( {\frac{D}{r}} \right)$$

Important Points:

For lossless lines, the three quantities Z, L, and C are related as:

$$h = \mu \frac{Z}{\eta },\;\;C =\epsilon \frac{\eta }{Z}$$

where,

η = Intrinsic Impedance of the dielectric medium

Putting the value of Z from equation (1) in L and C, we get:

$$L = \frac{\mu }{\eta }\eta \frac{h}{w} = \frac{{\mu h}}{w}$$

$$L = \frac{{\mu h}}{w}$$

L represents the inductance per unit length of parallel plate lines.

$$C = \frac{{E\;\eta \;w}}{{\eta \;h}} = \frac{{E\;w}}{h}$$

$$C = \frac{{E\;w}}{h}$$

C represents the Capacitance per unit length of parallel plate lines.

Speed of propagation is given by

$$v = \frac{1}{{\sqrt {LC} }}$$

$$v = \frac{1}{{\sqrt {\frac{{\mu h}}{w}\frac{{E\;w}}{h}} }} = \frac{1}{{\sqrt {\mu E} }}$$

v does not depend on the spacing between plates, hence v will not change

# A rectangular waveguide of width w and height h has cut-off frequencies for TE10 and TE11 modes in the ratio 1 : 2. The aspect ratio w/h, rounded off to two decimal places, is __________.

## Parallel Plane Waveguide MCQ Question 5 Detailed Solution

Concept:

The cutoff frequency of TEmn mode is given by:

$${f_c} = \frac{c}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}}$$

Application:

for TE10 mode

$${f_{{c_{10}}}} = \frac{c}{2}\left( {\frac{1}{a}} \right)$$        ----(1)

for TE11 mode

$${f_{{c_{11}}}} = \frac{c}{2}\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}}$$        ----(2)

$$\frac{{eqn\left( 1 \right)}}{{eqn\left( 2 \right)}} = \frac{1}{2}$$

$$\frac{1}{2} = \frac{{\frac{c}{2}\left( {\frac{1}{a}} \right)}}{{\frac{c}{2}\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} }}$$

$$= \frac{1}{2} = \frac{{\frac{c}{{2a}}}}{{\frac{c}{2}\sqrt {\frac{{{a^2} + {b^2}}}{{ab}}} }} = \frac{b}{{\sqrt {{b^2} + {a^2}} }}$$

$$\frac{1}{2} = \frac{1}{{\sqrt {1 + {{\left( {\frac{a}{b}} \right)}^2}} }}$$

$${\left( {\frac{a}{b}} \right)^2} = {\left( {\frac{w}{h}} \right)^2} = 4 - 1 = 3$$

$$\frac{w}{h} = \sqrt 3 = 1.732$$

# A rectangle waveguide of internal dimensions (a = 3 cm and b = 1 cm) is to be operated at TE11 mode. The minimum operating frequency is:

1. 6.25 GHz
2. 10.5 GHz
3. 31.6 GHz
4. 15.8 GHz

## Answer (Detailed Solution Below)

Option 4 : 15.8 GHz

## Parallel Plane Waveguide MCQ Question 6 Detailed Solution

Concept:

Minimum operating frequency or the cut off frequency for a rectangular waveguide is given by:

$${f_c} = \frac{c}{2}\sqrt {\frac{{{m^2}}}{{{a^2}}} + \frac{{{n^2}}}{{{b^2}}}}$$

a = length of the waveguide

b = height of the waveguide

m,n = modes of operation

Calculation:

Given, a = 3 cm

b = 1 cm

The minimum frequency in TE11 is nothing but the cut-off frequency calculated as:

$${f_c} = \frac{{3 \times {{10}^8}}}{2}\sqrt {\frac{1}{{{{\left( {3 \times {{10}^{ - 2}}} \right)}^2}}} + \frac{1}{{{{\left( {1 \times {{10}^{ - 2}}} \right)}^2}}}}$$

$${f_c} = 150 \times {10^6}\sqrt {11111.11} \;\;$$

$${f_c} = 150 \times {10^6} \times 105.41$$

fC = 15.81 GHz

# As the wave frequency approaches to cut-off frequency of wave guide, the correct statement is:

1. The phase velocity of waves tends to zero.
2. The phase velocity of waves tends to velocity of light.
3. The phase velocity of waves tends to infinite.
4. None of these.

## Answer (Detailed Solution Below)

Option 3 : The phase velocity of waves tends to infinite.

## Parallel Plane Waveguide MCQ Question 7 Detailed Solution

Concept:

Phase velocity:

$${{v}_{p}}=\frac{c}{\sqrt{1-{{\left( \frac{{{f}_{c}}}{f} \right)}^{2}}~}}$$

Where c is the free space velocity, f is the operating frequency and fc is the cut-off frequency.

1. We can readily see that the phase velocity is a non-linear function of the operating frequency.

2. We can also write phase velocity as:

$${v_p} = \frac{c}{{\cos \theta }};$$

Where θ is the angle with which the wave enters the waveguide as shown:

Calculation:

Given, f = fc

$${v_p} = \frac{c}{{\sqrt {1 - {{\left( {\frac{{{f_c}}}{f}} \right)}^2}} }}$$

$$v_p = \frac{c}{{\sqrt {1 - 1} \;}} = \frac{c}{0} = \infty$$

vp = ∞ means that,

cos θ = 0, i.e. θ = 90°.

This situation is shown as below:

This implies that at f = fc, the wave will oscillate between the walls as shown.

Hence option (3) is the correct answer.

# Transverse Electro-magnetic waves are characterized by

1. During wave propagation in Z-direction, the components of H and E are transverse 60° to the direction of propagation of the waves
2. During wave propagation in Z-direction, the components of H and E are transverse to the direction of propagation of the waves.
3. During wave propagation in Z-direction, the components of H and E are transverse 120° to the direction of propagation of the waves
4. None of the above

## Answer (Detailed Solution Below)

Option 2 : During wave propagation in Z-direction, the components of H and E are transverse to the direction of propagation of the waves.

## Parallel Plane Waveguide MCQ Question 8 Detailed Solution

Transverse Electro-magnetic waves:

• In an electromagnetic wave, electric and magnetic field vectors are perpendicular to each other and at the same time are perpendicular to the direction of propagation of the wave.
• This nature of electromagnetic wave is known as Transverse nature.
• Maxwell proved that both the electric(E) and magnetic fields(H) are perpendicular to each other in the direction of wave propagation. He considered an electromagnetic wave propagating along positive z-axis
• The electric and magnetic fields propagate sinusoidal with the z-axis.

# Standard air-filled rectangular waveguides of dimensions a = 2.29 cm and b = 1.02 cm are designed for radar applications. It is desired that these waveguides operate only in the dominant TE10 mode with the operating frequency at least 25% above the cutoff frequency of the TE10 mode but not higher than 95% of the next higher cutoff frequency. The range of the allowable operating frequency f is

1. 8.19 GHz ≤ f ≤ 13.1 GHz
2. 8.19 GHz ≤ f ≤ 12.45 GHz
3. 6.55 GHz ≤ f ≤ 13.1 GHz
4. 1.64 GHz ≤ f ≤ 10.24 GHz

## Answer (Detailed Solution Below)

Option 2 : 8.19 GHz ≤ f ≤ 12.45 GHz

## Parallel Plane Waveguide MCQ Question 9 Detailed Solution

Concept:

Cut-off frequency of waveguide (rectangular)

$$\Rightarrow {f_c} = \frac{C}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}} \;\;\;\; \ldots \left( 1 \right)$$

Dominant mode in rectangular waveguide (a > b) is TE10 and the cutoff frequency is defined as;

$${f_{{c_{\left( {10} \right)}}}} = \frac{C}{2}\sqrt {\frac{1}{{{a^2}}}} = \frac{C}{{2a}}$$

Calculation:

$${f_{{c_{\left( {10} \right)}}}} = \frac{C}{{2a}} = \frac{{3 \times {{10}^{10}}}}{{2 \times 2.29}} = 6.554\;Hz$$

Desired operating frequency = 25% more than $${f_{{c_{\left( {10} \right)}}}}$$

$$\left( {1 + 0.25} \right){f_{{c_{\left( {10} \right)}}}} = 1.25{f_{{c_{\left( {10} \right)}}}} = 1.25 \times 6.55$$

= 8.199 Hz

Using equation (1), we calculate different cut-off frequency for higher modes to find the next dominant mode,

fc for TE01 mode = 14.74 Hz

fc for TE11 mode = 16.4 Hz

and

fc for TE20 mode = 13.144 Hz

So, TE20 is the next higher cut-off frequency,

95% of 13.144 Hz

⇒ 12.484 GHz

# A standard air-filled rectangular waveguide with dimensions a = 8 cm, b = 4cm, operates at 3.4 GHz, For the dominant mode of wave propagation, the phase velocity of the signal is vp. The value (rounded off two decimal places) of vp / C, where C denotes the velocity of light, is _____

## Parallel Plane Waveguide MCQ Question 10 Detailed Solution

Concept:

Phase velocity is given by:

$${v_p} = \frac{\omega }{\beta }$$

Where, ω = angular frequency

β = phase constant, calculated for a parallel waveguide as:

$$\beta = \sqrt {{\omega ^2}\mu - {{\left( {\frac{{m\pi }}{a}} \right)}^2}}$$         [for parallel waveguide]

The Phase velocity can now be written as:

$${v_p} = \frac{\omega }{{\sqrt {{\omega ^2}\mu - {{\left( {\frac{{m\pi }}{a}} \right)}^2}} }}$$

$${v_p} = \frac{1}{{\sqrt {\mu - {{\left( {\frac{{m\pi }}{{a\omega }}} \right)}^2}} }}$$

$${v_p} = \frac{{\frac{1}{{\sqrt {\mu } }}}}{{\sqrt {1 - {{\left( {\frac{{m\pi }}{{a\omega \sqrt {\mu } }}} \right)}^2}} }}$$

With $$c = \frac{1}{{\sqrt {\mu } }}$$

$${v_p} = \frac{c}{{\sqrt {1 - {{\left( {\frac{{m\pi c}}{{a\omega }}} \right)}^2}} }}\;$$

Also, the cut-off frequency is defined as:

$${\omega _c} = \frac{{m{\pi _c}}}{a}$$

$${v_p} = \frac{c}{{\sqrt {1 - {{\left( {\frac{{{\omega _c}}}{\omega }} \right)}^2}} }}$$

$$\sin \theta = \frac{{{\omega _c}}}{\omega }$$

$${v_p} = \frac{c}{{\sqrt {1 - {{\sin }^2}\theta } }}$$

$${v_p} = \frac{C}{{\cos \theta }}$$

$${v_p} = \frac{c}{{\cos \theta }}$$

Note: This is valid for the Rectangular waveguide also.

Calculation:

Given

a = 8 cm

b = 4 cm

f = 3.4 GHz

The cut-off frequency will be:

$${f_c} = \frac{c}{{2a}} = \frac{{3 \times {{10}^{10}}}}{{2 \times 8}}$$

$$\sin \theta = \frac{{{f_c}}}{f}$$

$$\sin \theta = \frac{{1.875}}{{3.4}} = 0.5514$$

∴ cos θ = 0.83419

$$\frac{{{v_p}}}{c} = \frac{1}{{\cos \theta }}$$

$$\frac{{{v_p}}}{c} = 1.198$$

# The phase velocity of waves propagating in a hollow metal waveguide is

1. equal to the group velocity
2. equal to the velocity of light in free space
3. less than the velocity of light in free space
4. greater than the velocity of light in free space

## Answer (Detailed Solution Below)

Option 4 : greater than the velocity of light in free space

## Parallel Plane Waveguide MCQ Question 11 Detailed Solution

Concept:

Phase velocity is defined as:

$${V_p} = \frac{\omega }{\beta }$$

β is the phase constant defined as:

$$\beta = \sqrt {{\omega ^2}\mu \epsilon - {{\left( {\frac{{m\pi }}{a}} \right)}^2}}$$

$${V_p} = \frac{\omega }{{\sqrt {{\omega ^2}\mu \epsilon - {{\left( {\frac{{m\pi }}{a}} \right)}^2}} }}$$

$${V_p} = \frac{1}{{\sqrt {\mu \epsilon - {{\left( {\frac{{m\pi }}{{a\omega }}} \right)}^2}} }}$$

$${V_p} = \frac{{\frac{1}{{\sqrt {\mu \epsilon } }}}}{{\sqrt {1 - {{\left( {\frac{{m\pi }}{{a\omega \sqrt {\mu \epsilon} }}} \right)}^2}} }}$$

Using $$c = \frac{1}{{\sqrt {\mu C} }}$$ where c =  speed of light, the above expression becomes:

$${V_p} = \frac{c}{{\sqrt {1 - {{\left( {\frac{{m\pi C}}{{a\omega }}} \right)}^2}} }};$$

Also $${\omega _c} = \frac{{m\pi c}}{a}$$

$${V_p} = \frac{C}{{\sqrt {1 - {{\left( {\frac{{{\omega _c}}}{\omega }} \right)}^2}} }}$$

Using $$\sin \theta = \frac{{{\omega _c}}}{\omega }$$, we get:

$${V_p} = \frac{c}{{\sqrt {1 - {{\sin }^2}\theta } }}$$

$${V_p} = \frac{c}{{\cos \theta }};$$

Since -1 ≤ cos θ ≤ 1

Vp > c

Extra Information:

Group velocity is given by:

$${V_g} = \frac{{d\omega }}{{d\beta }}$$

$$\beta = \sqrt {{\omega ^2}\mu \epsilon - {{\left( {\frac{{m\pi }}{a}} \right)}^2}}$$

$$\frac{{d\beta }}{{d\omega }} = \frac{{2\omega \mu\epsilon }}{{2\sqrt {{\omega ^2}\mu\epsilon - {{\left( {\frac{{m\pi }}{a}} \right)}^2}} }}$$

$$\frac{{d\beta }}{{d\omega }} = \frac{{\sqrt {\mu\epsilon } }}{{\sqrt {1 - {{\left( {\frac{{m\pi }}{{a\omega \sqrt {\mu\epsilon } }}} \right)}^2}} }}$$

$$\frac{{d\beta }}{{d\omega }} = \frac{1}{{C\sqrt {1 - {{\left( {\frac{{{\omega _C}}}{\omega }} \right)}^2}} }}$$

$${V_g} = c\sqrt {1 - {{\left( {\frac{{{\omega _C}}}{\omega }} \right)}^2}}$$

Vg = c cos θ

Vg < c

Conclusion: ∴ The phase velocity is always greater than the speed of light and group velocity is always less than the speed of light.

# For transverse electric waves between parallel plates, the lowest value of m, without making all the field components zero, is equal to

1. 3
2. 2
3. 1
4. 0

Option 3 : 1

## Parallel Plane Waveguide MCQ Question 12 Detailed Solution

Concept:

For parallel plate waveguide, the propagation constant is defined as:

$$\beta = \frac{{m\pi }}{a}$$

a = distance between the plates

Where the minimum value of m = 1

Cut-off frequency for the parallel plate waveguide is given by:

$${f_c} = \frac{m}{{2a}}$$

The wavelength will be:

$${\lambda _c} = \frac{{2a}}{m}$$

# The cut-off frequency of TE01 mode of an air-filled rectangular waveguide having inner dimensions a cm × b cm (a > b) is twice that of the dominant TE10 mode. When the waveguide is operated at a frequency which is 25% higher than the cut-off frequency of the dominant mode, the guide wavelength is found to be 4 cm. The value of b (in cm, correct to two decimal places) is _______.

## Parallel Plane Waveguide MCQ Question 13 Detailed Solution

Concept:

Dominant mode of rectangular waveguide is TE10

The cutoff frequency is given as:

$${f_c} = \frac{c}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}}$$

Calculations:

Given: $${\left( {{f_c}} \right)_{TE01}} = 2{\left( {f_c} \right)_{TE10}}$$     ---(1)

Operating frequency:

f = (fc)TE10 + 25% of (fc)TE10

f = 1.25(fc)TE10

For air filled wavelength C, we can write:

$$f_c = \frac{c}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}}$$

For TE01 mode

m = 0, n = 1

$${\left( {fc} \right)_{TE10}} = \frac{c}{{2a}}$$

For TE10 mode

m = 1; n = 0

$${\left( {f_c} \right)_{TE10}} = \frac{c}{{2a}}$$     ----(2)

From equation (1)

(fc)TE01 = 2(fc)TE10

$$\frac{c}{{2b}} = 2.\frac{c}{{2a}}$$

a = 2b

operating frequency:

f = (1.25) (fc)TE01 = 2(fc)TE10

$$f = \left( {1.25} \right)\left[ {\frac{c}{{2a}}} \right]$$

$$= \frac{{1.25\;c}}{{4b}}$$

Operating wavelength

$$\lambda = \frac{c}{f} = \frac{{4b}}{{1.25}} = 3.2b$$

Guide wavelength is given by:

$${\lambda _g} = \frac{\lambda }{{\sqrt {1 - {{\left( {\frac{{{f_c}}}{f}} \right)}^2}} }}$$

$$4 = \frac{{3.2b}}{{\sqrt {1 - {{\left( {\frac{1}{{1.25}}} \right)}^2}} }}$$

$$4 = \frac{{3.2b}}{{0.6}}$$

$$\Rightarrow b = \frac{{4 \times 0.6}}{{3.2}}$$

b = 0.75 cm
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