Home
Class 12
PHYSICS
A microscope has objective of aperture 8...

A microscope has objective of aperture 8mm and focal length `2.5 cm`. Estimate its resolving power. Given `lambda= 5500 dotA`.

Text Solution

AI Generated Solution

To estimate the resolving power of the microscope, we can follow these steps: ### Step 1: Understand the given values We have the following information: - Aperture (D) = 8 mm = 8 × 10^-3 m - Focal length (f) = 2.5 cm = 2.5 × 10^-2 m - Wavelength (λ) = 5500 Å = 5500 × 10^-10 m = 5.5 × 10^-7 m ...
Promotional Banner

Topper's Solved these Questions

  • WAVES OPTICS

    AAKASH SERIES|Exercise EXERCISE -IA|54 Videos

  • WAVES OPTICS

    AAKASH SERIES|Exercise EXERCISE -IA (DIFFRACTION)|33 Videos

  • WAVES

    AAKASH SERIES|Exercise EXERCISE-III (Doppler effect :)|15 Videos

Similar Questions

Explore conceptually related problems

A compound microscope has an objective of focal length 1 cm and an eyepiece of focal length 2.5 cm. An object has to be placed at a distance of 1.2 cm away from the objective for normal adjustment. a.Find the angular magnification. b.Find the length of the microscope tube.

A person with a normal near point (25 cm) using a compound microscope with an objective of focal length 8.0 mm and eye piece of focal length 2.5 cm can bring an object placed 9.0 cm from the objective in sharp focus. What is the separation between the two lenses ? Calculate the magnifying power of the microscope ?

A person with a normal near point (25 cm) using a compound microscope with an objective of focal length 8.0 mm and eye piece of focal length 2.5 cm can bring an object placed 9.0 cm from the objective in sharp focus. What is the separation between the two lenses ? Calculate the magnifying power of the microscope ?

A telescope has an objective of focal length 100 cm and an eye-piece of focal length 5 cm. What is the magnifying power of the telescope when it is in normal adjustment?

A compound microscope has objective of focal length 3.2 mm and eyepiece of 12 mm focal length. If the objective forms image of the object 16 cm beyond its pole, then total magnification achieved is

A compound microscope consists of an objective of focal length 1 cm and an eyepiece of focal length 5 cm. An object is placed at a distance of 0.5 cm from the objective. What should be the separation between the lenses so that the microscope projects an inverted real image of the object on a screen 30 cm behind the eyepiece?

A compound microscope has an objective of focal length 2.0 cm and an eye-piece of focal length 6.25cm and distance between the objective and eye-piece is 15cm . If the final image is formed at the least distance vision (25 cm) , the distance of the object form the objective is

A small telescope has an objective lens of focal length 144 cm and an eye-piece of focal length 6.0 cm . What is the magnifying power of the telescope ? What is the separation between the objective and the eye-piece ?

A small telescope has an objective lens of focal length 144 cm and an eye-piece of focal length 6.0 cm . What is the magnifying power of the telescope ? What is the separation between the objective and the eye-piece ?

A compound microscope uses an objective lens of focal length 4 cm and eye lens of focal length 10 cm . An object is placed at 6 cm from the objective lens. Calculate magnifying power of compound microscope if final image is formed at the near point. Also, calculate length of the tube of compound microscope.