Air France AF4107 Flight Status Today

International flight Air France AF4107 takes off from Mauritius (MRU) Mauritius to Paris (CDG) France. It's operated by Air France. The plane leaves Sir Seewoosagur Ramgoolam International Airport at 10:45 Indian/Mauritius. The flight is expected to land at Charles de Gaulle Airport at 20:25 Europe/Paris. The flight will last about 11 hours 48 minutes.

AF4107 - AFR4107 Air France (AF)
MRU Mauritius Mauritius
Flight MRU to CDG
CDG Paris France
Contact Airlines It may not operate on the date requested
Departure Sir Seewoosagur Ramgoolam International Airport
IATA: MRUICAO: FIMP
Scheduled 10:45
None None
Arrival Charles de Gaulle Airport
IATA: CDGICAO: LFPG
Scheduled 20:25
2E None
AF4107 Detail
  • Type: International Flight
  • Flight Duration: 11 hours 48 minutes
  • Flight Distance: 9425 kms / 5856 miles
Airline
  • Air France
  • IATA: AF
  • ICAO: AFR
  • Operating Days: Monday
Passenger Services
  • Service type: None
  • Seats: None
  • Freight capacity: None
  • Passenger classes: None
More Detail
  • Aircraft: Boeing 777-300ER
  • Callsign: None
  • Departure Timezone: Indian/Mauritius
  • Arrival Timezone: Europe/Paris
  • Current Time in Mauritius: Monday 2026-04-13 03:42 AM
  • Current Time in Paris: Monday 2026-04-13 01:42 AM

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Air France Flight AF4107 FAQs

What is the scheduled flight duration for Air France AF4107 flight?

On average, nonstop flight takes 11 hour(s) 48 minutes, with the flight distance of 9425 km (5856 miles).

What type of aircraft is used for the Air France AF4107 flight?

All Air France AF4107 flights are operated using Boeing 777-300ER aircraft.

Which terminal the flight Air France AF4107 is arriving at?

Flight Air France AF4107 arrives in Charles de Gaulle Airport at Terminal 2E.

How many Air France AF4107 flights are operated a week?

1 flights per week. The Flight Air France AF4107 is operated on Monday.