Optical Communication MCQs (Multiple-Choice Questions)

1. What is the primary purpose of an optical communication system?

1. To transmit information using light signals
2. To transmit information only through sound waves
3. To store electrical energy in batteries
4. To convert all electrical signals into magnetic fields

Answer: A) To transmit information using light signals

Explanation:

An optical communication system transmits information by using light as the carrier. In fiber-optic systems, the light travels through an optical fiber and carries information encoded through techniques such as intensity, phase, or frequency modulation.

2. Which medium is commonly used to guide light in a fiber-optic communication system?

1. Copper wire
2. Optical fiber
3. Steel cable
4. Coaxial conductor only

Answer: B) Optical fiber

Explanation:

An optical fiber is a dielectric waveguide, typically made from glass or plastic, that guides light along its length. Glass fibers are widely used for high-capacity telecommunications because of their low attenuation at suitable operating wavelengths.

3. Which phenomenon is primarily responsible for guiding light in a conventional step-index optical fiber?

1. Photoelectric emission
2. Magnetic induction
3. Total internal reflection
4. Electrolysis

Answer: C) Total internal reflection

Explanation:

In a conventional step-index fiber, the core has a higher refractive index than the surrounding cladding. Light traveling within the core can be confined by total internal reflection when the relevant incidence-angle conditions are satisfied. More generally, optical fibers guide light through supported electromagnetic modes.

4. What are the two main optical layers of a conventional optical fiber?

1. Anode and cathode
2. Transmitter and receiver
3. Insulator and conductor
4. Core and cladding

Answer: D) Core and cladding

Explanation:

The core is the central region through which the optical signal is guided, while the cladding surrounds the core and has a lower refractive index in a conventional step-index fiber. A protective coating is usually applied outside the cladding to protect the fiber from damage and moisture.

5. Which material property must be higher in the core than in the cladding for conventional step-index fiber guidance by total internal reflection?

1. Electrical conductivity
2. Refractive index
3. Magnetic permeability in every case
4. Electrical resistance

Answer: B) Refractive index

Explanation:

For conventional step-index fiber guidance by total internal reflection, the core refractive index must be greater than the cladding refractive index. This index contrast permits light in appropriate guided modes to remain confined to the core.

6. Which of the following is a common optical source used in fiber-optic communication?

1. Laser diode
2. Transformer
3. Relay switch
4. Thermistor

Answer: A) Laser diode

Explanation:

A laser diode converts electrical energy into optical radiation and is widely used in fiber-optic transmitters. Its wavelength, output power, modulation response, and spectral characteristics depend on the device design and application.

7. Which optical source is commonly used in short-distance, relatively low-cost fiber-optic links?

1. Microwave antenna
2. Phototransistor used as a transmitter
3. Light-emitting diode (LED)
4. Piezoelectric sensor

Answer: C) Light-emitting diode (LED)

Explanation:

LEDs are used in some short-distance optical links because they are relatively simple and economical. Compared with many laser diodes, LEDs generally have lower optical output power, a broader spectral width, and lower modulation bandwidth.

8. What is the main function of an optical transmitter?

1. To amplify sound signals
2. To measure the physical length of the fiber only
3. To convert received light into an electrical signal
4. To convert the information signal into a suitable optical signal

Answer: D) To convert the information signal into a suitable optical signal

Explanation:

An optical transmitter typically includes electronic drive circuitry and an optical source, such as an LED or laser diode. It converts the input information into an optical signal that can be coupled into the transmission fiber.

9. Which component converts an incoming optical signal into an electrical signal at the receiver?

1. Optical isolator
2. Photodetector
3. Optical connector
4. Fiber cladding

Answer: B) Photodetector

Explanation:

A photodetector converts incident optical power into an electrical signal. Common devices used in optical communication receivers include PIN photodiodes and avalanche photodiodes (APDs).

10. Which photodetector is widely used in optical fiber communication receivers?

1. Thermocouple
2. Strain gauge
3. PIN photodiode
4. Induction coil

Answer: C) PIN photodiode

Explanation:

A PIN photodiode is a semiconductor photodetector with p-type, intrinsic, and n-type regions. It converts received photons into charge carriers, producing a photocurrent related to the incident optical power within its operating range.

11. What is attenuation in an optical fiber?

1. A reduction in optical signal power as light propagates through the fiber
2. An increase in the physical diameter of the fiber
3. A change in electrical resistance only
4. A conversion of light into sound at the transmitter

Answer: A) A reduction in optical signal power as light propagates through the fiber

Explanation:

Attenuation describes the reduction in optical power during transmission. Absorption, scattering, bending losses, and connector or splice losses can contribute to signal loss. Fiber attenuation is commonly expressed in decibels per kilometer (dB/km).

12. Which unit is commonly used to express optical fiber attenuation per unit length?

1. Watt per second
2. Hertz per meter
3. Joule per kilometer
4. Decibel per kilometer (dB/km)

Answer: D) Decibel per kilometer (dB/km)

Explanation:

Fiber attenuation is commonly specified in dB/km. If the attenuation coefficient is known, the approximate fiber loss in decibels can be calculated by multiplying the attenuation coefficient by the fiber length, assuming the coefficient is uniform over the link.

13. What is modal dispersion in a multimode optical fiber?

1. A reduction in the fiber's physical diameter
2. A spreading of an optical pulse because different modes travel with different propagation delays
3. A conversion of optical power into electrical power at the transmitter
4. A complete elimination of signal noise

Answer: B) A spreading of an optical pulse because different modes travel with different propagation delays

Explanation:

In multimode fibers, different guided modes can arrive at different times because they follow different propagation paths or have different group delays. This pulse spreading can cause adjacent digital pulses to overlap and limit the transmission bandwidth-distance product.

14. Which type of optical fiber generally has the largest core diameter?

1. Single-mode fiber
2. Hollow-core fiber in every design
3. Multimode fiber
4. All optical fibers have identical core diameters

Answer: C) Multimode fiber

Explanation:

Conventional multimode fibers have larger cores than standard single-mode telecommunications fibers. Common multimode core diameters include 50 µm and 62.5 µm, while standard single-mode fiber cores are typically around 8–10 µm in diameter.

15. Which optical fiber is generally preferred for long-distance, high-capacity telecommunications?

1. Single-mode fiber
2. Plastic optical fiber in every application
3. Multimode step-index fiber only
4. Fiber with an electrically conductive core

Answer: A) Single-mode fiber

Explanation:

Single-mode fiber supports one fundamental spatial mode under normal operating conditions. It avoids intermodal dispersion, making it well suited to long-distance and high-capacity links. Chromatic dispersion and other effects still need to be managed.

16. What is the main advantage of single-mode optical fiber over multimode optical fiber for long-distance transmission?

1. It always has a larger core
2. It eliminates every source of attenuation
3. It requires no optical transmitter
4. It avoids intermodal dispersion

Answer: D) It avoids intermodal dispersion

Explanation:

Single-mode fiber supports only one fundamental spatial mode at its operating wavelength, so there is no intermodal pulse spreading. However, chromatic dispersion, polarization-mode dispersion, attenuation, and nonlinear effects can still affect performance.

17. What is numerical aperture (NA) in an optical fiber?

1. A measure of electrical conductivity
2. A measure of the fiber's light-gathering ability related to its acceptance angle
3. The number of electrical connectors on a fiber
4. A measure of optical power loss in dB/km

Answer: B) A measure of the fiber's light-gathering ability related to its acceptance angle

Explanation:

Numerical aperture describes the acceptance capability of an optical fiber. For a step-index fiber in air, under the usual approximation, NA = √(n_core² − n_cladding²), where n_core and n_cladding are the refractive indices of the core and cladding. A higher NA generally corresponds to a larger acceptance angle.

18. What is the critical angle for total internal reflection at a core-cladding interface?

1. The angle measured from the fiber axis at which all light is absorbed
2. An angle that depends only on fiber length
3. The incidence angle, measured from the normal, for which the refracted ray travels along the interface
4. The angle at which light changes into an electrical current

Answer: C) The incidence angle, measured from the normal, for which the refracted ray travels along the interface

Explanation:

When light travels from a medium with a higher refractive index to one with a lower refractive index, the critical angle is the incidence angle for which the refracted angle is 90 degrees. For angles greater than the critical angle, total internal reflection occurs in the ideal ray model.

19. Which wavelength is commonly used in long-distance silica fiber-optic communication systems?

1. 850 nm only
2. 10 nm
3. 100 nm
4. 1550 nm

Answer: D) 1550 nm

Explanation:

The wavelength region around 1550 nm is widely used in long-distance silica fiber links because standard telecommunications fiber has low attenuation in this region, and optical amplifiers such as erbium-doped fiber amplifiers can operate there. Other wavelengths, including around 1310 nm and 850 nm, are used in suitable applications.

20. Why is the wavelength region around 1550 nm important in optical communication?

1. It is a widely used low-attenuation region in silica fiber and supports optical amplification with suitable amplifiers
2. Silica fiber cannot transmit any light at this wavelength
3. It completely eliminates all forms of dispersion
4. It does not require a receiver

Answer: A) It is a widely used low-attenuation region in silica fiber and supports optical amplification with suitable amplifiers

Explanation:

Standard silica fiber has a low-attenuation window around 1550 nm. Erbium-doped fiber amplifiers can amplify signals in a band around this wavelength, making it useful for long-distance links. The wavelength does not eliminate dispersion or other impairments.

21. What is chromatic dispersion in an optical fiber?

1. A change in fiber diameter due to pressure
2. Pulse spreading caused by different wavelength components traveling at different group velocities
3. A complete loss of the optical signal at the transmitter
4. A method of increasing electrical current in a receiver

Answer: B) Pulse spreading caused by different wavelength components traveling at different group velocities

Explanation:

Chromatic dispersion occurs because different spectral components of an optical signal can propagate with different group velocities. The resulting pulse spreading can limit data rates and transmission distances. It includes material and waveguide contributions.

22. Which type of fiber has a refractive index that decreases gradually from the center of the core toward the cladding?

1. Metallic fiber
2. Uniform-index copper cable
3. Graded-index fiber
4. Coaxial electrical cable

Answer: C) Graded-index fiber

Explanation:

In graded-index fiber, the core refractive index varies gradually with radial position. A suitable index profile reduces modal dispersion by compensating for differences in the propagation times of modes traveling along different paths.

23. What is the purpose of cladding in an optical fiber?

1. To provide the refractive-index boundary that helps confine guided light and protect the core
2. To convert the optical signal into an electrical signal
3. To generate the information signal at the transmitter
4. To increase electrical conductivity in the core

Answer: A) To provide the refractive-index boundary that helps confine guided light and protect the core

Explanation:

The cladding surrounds the core and, in conventional fiber, has a lower refractive index. This index contrast enables total internal reflection for appropriate guided rays and supports optical mode confinement. The cladding also separates the core from the protective coating.

24. Which phenomenon is primarily responsible for intrinsic scattering loss in silica optical fiber?

1. Electrical conduction through the core
2. Mechanical rotation of the transmitter
3. Sound-wave reflection at the receiver
4. Rayleigh scattering caused by microscopic refractive-index fluctuations

Answer: D) Rayleigh scattering caused by microscopic refractive-index fluctuations

Explanation:

Rayleigh scattering occurs because microscopic density and refractive-index fluctuations in glass scatter a small portion of the optical power. It is an important intrinsic attenuation mechanism in silica fibers and generally becomes weaker at longer wavelengths.

25. What is the primary function of an optical amplifier in a fiber-optic communication link?

1. To convert every optical signal into sound
2. To increase the optical signal power without first converting the signal to an electrical one
3. To eliminate the need for optical fiber
4. To reduce the transmitter wavelength to zero

Answer: B) To increase the optical signal power without first converting the signal to an electrical one

Explanation:

An optical amplifier increases optical signal power directly in the optical domain. Examples include erbium-doped fiber amplifiers and semiconductor optical amplifiers. Amplification also introduces noise and does not automatically correct accumulated dispersion or distortion.

26. What does WDM stand for in optical communication?

1. Waveform Data Measurement
2. Wireless Distribution Method
3. Wavelength Division Multiplexing
4. Wideband Digital Modulation

Answer: C) Wavelength Division Multiplexing

Explanation:

Wavelength Division Multiplexing (WDM) transmits multiple optical channels at different wavelengths through the same fiber. Each wavelength carries a separate data stream, increasing the capacity of the fiber without requiring a separate fiber for every channel.

27. What is the main advantage of Wavelength Division Multiplexing (WDM)?

1. It allows multiple optical channels to share the same fiber using different wavelengths
2. It eliminates the need for receivers
3. It converts optical fibers into electrical conductors
4. It guarantees zero attenuation over any distance

Answer: A) It allows multiple optical channels to share the same fiber using different wavelengths

Explanation:

WDM increases fiber capacity by carrying multiple wavelength channels simultaneously. Each channel can carry its own information stream. The actual capacity depends on factors such as channel spacing, modulation format, signal-to-noise ratio, and fiber impairments.

28. Which technique combines several lower-rate digital signals into a higher-rate aggregate signal in a communication system?

1. Attenuation
2. Rayleigh scattering
3. Optical absorption
4. Time Division Multiplexing (TDM)

Answer: D) Time Division Multiplexing (TDM)

Explanation:

Time Division Multiplexing allocates different time slots to separate signals on a shared transmission resource. In optical systems, TDM can be used in the electronic or optical domain, while WDM separates channels by wavelength.

29. What is the main function of an optical connector?

1. To generate laser light
2. To provide a detachable optical connection between fibers or optical devices
3. To measure the speed of light in a vacuum
4. To convert digital data into sound

Answer: B) To provide a detachable optical connection between fibers or optical devices

Explanation:

An optical connector allows a fiber to be connected to another fiber or to an optical device. Proper alignment and clean end faces help minimize insertion loss and reflections at the connection.

30. What is insertion loss in an optical communication link?

1. The increase in fiber core diameter after installation
2. The optical signal's frequency in hertz
3. The optical power loss introduced by inserting a component or connection into the link
4. The total number of transmitted bits

Answer: C) The optical power loss introduced by inserting a component or connection into the link

Explanation:

Insertion loss measures the reduction in transmitted optical power caused by inserting a component, such as a connector, splice, filter, or coupler. It is usually expressed in decibels and is an important parameter in optical link budgeting.

31. Which instrument is commonly used to measure optical power in a fiber-optic system?

1. Optical power meter
2. Hydrometer
3. Barometer
4. Seismograph

Answer: A) Optical power meter

Explanation:

An optical power meter measures the optical power received at a specified wavelength. When used with a suitable optical source, it can also help determine insertion loss or fiber attenuation over a test link.

32. What is the purpose of an Optical Time-Domain Reflectometer (OTDR)?

1. To measure electrical resistance in a copper cable only
2. To generate digital data at a transmitter
3. To increase the refractive index of the fiber
4. To locate fiber events and estimate loss by analyzing backscattered light and reflections

Answer: D) To locate fiber events and estimate loss by analyzing backscattered light and reflections

Explanation:

An OTDR sends optical pulses into a fiber and analyzes the light returned by backscattering and reflections. The resulting trace can help locate breaks, connectors, splices, and other events, as well as estimate fiber attenuation and event losses.

33. What is responsivity in a photodetector?

1. The physical length of the detector
2. The ratio of output photocurrent to incident optical power under specified conditions
3. The number of connectors in a receiver
4. The ratio of fiber length to core diameter

Answer: B) The ratio of output photocurrent to incident optical power under specified conditions

Explanation:

Photodetector responsivity is commonly expressed in amperes per watt (A/W). It describes the photocurrent produced per unit of incident optical power at a specified wavelength and operating condition.

34. Which statement about an avalanche photodiode (APD) is correct?

1. It operates only as an optical transmitter
2. It cannot detect photons
3. It uses internal avalanche multiplication to provide photodetector gain
4. It eliminates all noise in an optical receiver

Answer: C) It uses internal avalanche multiplication to provide photodetector gain

Explanation:

An APD operates with a sufficiently high reverse bias to produce impact ionization and internal carrier multiplication. This gain can improve sensitivity in some receivers, but it also introduces excess multiplication noise and requires careful bias control.

35. Which factor can cause bending loss in an optical fiber?

1. Bending the fiber more sharply than its specified bend radius
2. Using a clean optical connector
3. Measuring the optical power at the correct wavelength
4. Using a suitable optical receiver

Answer: A) Bending the fiber more sharply than its specified bend radius

Explanation:

When an optical fiber is bent too sharply, some guided optical power may escape from the core. Tight bends can cause macrobending loss, while small-scale irregularities can contribute to microbending loss. Following the fiber's bend-radius specifications helps reduce these losses.

36. What is the purpose of modulation in an optical communication system?

1. To increase the physical diameter of the optical fiber
2. To remove the need for an information source
3. To ensure that all optical signals have the same wavelength
4. To encode information onto an optical carrier by varying a suitable signal parameter

Answer: D) To encode information onto an optical carrier by varying a suitable signal parameter

Explanation:

Modulation encodes information onto an optical carrier. Depending on the system, the information may be represented through changes in optical intensity, phase, or frequency. The receiver detects and processes the modulated signal to recover the transmitted information.

37. What does bit error rate (BER) measure in a digital optical communication system?

1. The physical diameter of the fiber
2. The proportion of received bits that are detected incorrectly
3. The number of optical connectors installed
4. The total fiber length in kilometers

Answer: B) The proportion of received bits that are detected incorrectly

Explanation:

Bit error rate is the number of incorrectly detected bits divided by the total number of received bits observed over a measurement interval. BER is used to assess digital link performance and can be affected by noise, dispersion, interference, and receiver limitations.

38. Which factor can limit the maximum data rate of an optical fiber link?

1. Fiber color alone
2. The number of protective cable labels
3. Dispersion and the bandwidth of the transmitter and receiver
4. The physical color of the connector housing alone

Answer: C) Dispersion and the bandwidth of the transmitter and receiver

Explanation:

Dispersion can spread pulses and cause intersymbol interference at high data rates. The bandwidth and response characteristics of the optical source, receiver, and associated electronics also constrain transmission performance. Noise and nonlinear effects can impose additional limits.

39. What is the approximate speed of light in a medium with refractive index n?

1. v = c/n
2. v = cn
3. v = n/c
4. v = c + n

Answer: A) v = c/n

Explanation:

The phase velocity of light in a homogeneous, nonmagnetic dielectric is approximately v = c/n, where c is the speed of light in a vacuum and n is the refractive index. In fiber-optic pulse-propagation calculations, group velocity is often the more relevant quantity.

40. Which statement about optical fiber communication is correct?

1. Optical fibers are highly susceptible to electromagnetic interference in the same way as unshielded copper wires
2. Optical fibers cannot carry digital data
3. Optical fiber transmission always has zero attenuation
4. The optical transmission path is immune to electromagnetic interference because it carries light rather than electrical current

Answer: D) The optical transmission path is immune to electromagnetic interference because it carries light rather than electrical current

Explanation:

The optical signal in a dielectric fiber is not affected by electromagnetic interference in the same way as an electrical signal in a metallic conductor. However, electronic equipment at the transmitter and receiver can still be affected by electromagnetic interference, and the optical link remains subject to attenuation, dispersion, and noise.

41. An optical fiber has an attenuation coefficient of 0.2 dB/km. What is the approximate fiber attenuation over a length of 10 km, excluding connector and splice losses?

1. 0.2 dB
2. 2 dB
3. 10 dB
4. 20 dB

Answer: B) 2 dB

Explanation:

For a uniform fiber, fiber attenuation in decibels is approximately the attenuation coefficient multiplied by the length. Therefore, total fiber attenuation = 0.2 dB/km × 10 km = 2 dB. This calculation excludes losses from connectors, splices, and other components.

42. An optical signal has an input power of 10 mW and an output power of 1 mW. What is the optical power loss in decibels?

1. 1 dB
2. 3 dB
3. 10 dB
4. 20 dB

Answer: C) 10 dB

Explanation:

Optical power loss is calculated as L = 10 log10(P_in/P_out). Substituting P_in = 10 mW and P_out = 1 mW gives L = 10 log10(10) = 10 dB.

43. What is the numerical aperture of a step-index optical fiber in air if the core refractive index is 1.48 and the cladding refractive index is 1.46?

1. 0.243
2. 0.500
3. 1.460
4. 0.243 is approximately correct only if calculated from the refractive-index difference without the square root

Answer: A) 0.243

Explanation:

For a step-index fiber in air, NA = √(n_core² − n_cladding²). Substituting the values gives NA = √(1.48² − 1.46²) = √(2.1904 − 2.1316) = √0.0588 ≈ 0.243. The result is dimensionless.

44. A fiber-optic link has 15 km of fiber with attenuation of 0.25 dB/km. It also has total connector and splice losses of 1.5 dB. What is the total link loss?

1. 1.5 dB
2. 2.5 dB
3. 3.0 dB
4. 5.25 dB

Answer: D) 5.25 dB

Explanation:

Fiber loss = 15 km × 0.25 dB/km = 3.75 dB. Adding connector and splice losses gives total loss = 3.75 dB + 1.5 dB = 5.25 dB. This calculation does not include additional engineering margin or other unlisted losses.

45. Which statement best describes the relationship between optical power and power expressed in dBm?

1. dBm = 10 log10(P in mW)
2. dBm = 10 log10(P in watts)
3. dBm = P in milliwatts divided by 10
4. dBm = 20 log10(P in mW)

Answer: A) dBm = 10 log10(P in mW)

Explanation:

dBm expresses power relative to 1 milliwatt. The formula is P_dBm = 10 log10(P_mW), where P_mW is the power in milliwatts. For example, 1 mW equals 0 dBm, and 10 mW equals 10 dBm.

46. An optical receiver requires at least −25 dBm of received power. The transmitter output is 0 dBm, and the total link loss is 20 dB. What is the received power, and does the link meet the receiver's minimum power requirement?

1. −20 dBm; yes, with 5 dB of power margin
2. −25 dBm; no, because the received power is zero
3. +20 dBm; yes, with 45 dB of power margin
4. −5 dBm; no, because the received power is below the minimum

Answer: A) −20 dBm; yes, with 5 dB of power margin

Explanation:

Received power in dBm equals transmitter power in dBm minus total link loss in dB. Therefore, P_rx = 0 dBm − 20 dB = −20 dBm. Since −20 dBm is 5 dB higher than the minimum required power of −25 dBm, the link meets the stated power requirement with a 5 dB margin, assuming no other impairments or additional losses.

47. Which statement about an optical fiber's acceptance angle is correct for a step-index fiber in air?

1. The acceptance angle is always 90 degrees for every fiber
2. It depends on the fiber's numerical aperture and describes the maximum launch angle, measured from the fiber axis, for guided propagation in the ideal ray model
3. It is the angle between two electrical conductors
4. It is independent of the refractive indices of the core and cladding

Answer: B) It depends on the fiber's numerical aperture and describes the maximum launch angle, measured from the fiber axis, for guided propagation in the ideal ray model

Explanation:

For launch from air into a step-index fiber, the acceptance half-angle θ_max satisfies sin(θ_max) = NA when NA is no greater than 1. A higher numerical aperture generally permits a wider range of launch angles to couple into guided modes.

48. Which technique is commonly used to reduce intermodal dispersion in a multimode fiber?

1. Increasing connector contamination
2. Replacing the optical source with an electrical transformer
3. Using a graded-index refractive-index profile
4. Increasing the fiber's attenuation

Answer: C) Using a graded-index refractive-index profile

Explanation:

A graded-index profile changes the refractive index gradually across the core. This helps compensate for differences in propagation times among modes, reducing modal dispersion compared with a conventional multimode step-index fiber with a similar core size.

49. What is the main purpose of a link budget in optical communication?

1. To compare available transmitter power with expected link losses and receiver sensitivity
2. To calculate the physical color of the optical cable
3. To determine the number of electrons in a laser diode only
4. To eliminate all dispersion without changing the system design

Answer: A) To compare available transmitter power with expected link losses and receiver sensitivity

Explanation:

A link budget accounts for transmitter output power, fiber attenuation, connector and splice losses, other component losses, receiver sensitivity, and any required design margin. It helps determine whether sufficient optical power will reach the receiver under the planned operating conditions.

50. An optical communication link uses a transmitter with an output power of +3 dBm. The fiber is 20 km long and has attenuation of 0.2 dB/km. The link also has two connectors, each with an insertion loss of 0.5 dB, and one splice with a loss of 0.2 dB. Ignoring all other losses, what is the received optical power?

1. −3.7 dBm
2. −2.2 dBm
3. −1.7 dBm
4. −5.2 dBm

Answer: B) −2.2 dBm

Explanation:

Fiber attenuation = 20 km × 0.2 dB/km = 4 dB.

Connector loss = 2 × 0.5 dB = 1 dB.

Splice loss = 0.2 dB.

Total link loss = 4 + 1 + 0.2 = 5.2 dB.

Received optical power = Transmitter power − Total link loss = +3 dBm − 5.2 dB = −2.2 dBm.

Therefore, the received optical power is −2.2 dBm.