Question

Notify Me Bookmark

The object-lens distance for a certain converging lens is 300 mm. The image is three times the size of the object. To make the image five times the size of the object, the object-lens distance must be changed to 450 mm 405 mm 540 mm 720 mm 270 mm

The object-lens distance for a certain converging lens is 300 mm. The image is three times the size of the object. To make the image five times the size of the object, the object-lens distance must be changed to 450 mm 405 mm 540 mm 720 mm 270 mm

Image text
The object-lens distance for a certain converging lens is 300 mm . The image is three times the size of the object. To make the image five times the size of the object, the object-lens distance must be changed to 450 mm 405 mm 540 mm 720 mm 270 mm

Detailed Answer

3 0

Please enter your email here. You will get email when this question is answered by our tutor.

Answer
Doubtrix Logo

This Problem has been solved!

You'll get a detailed, step-by-step and expert verified solution.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Login Sign Up
  • Student Reviews:
  • (3)
  • Correct answers (3)
  • Complete solution (3)
  • Step-by-step solution (3)
  • Fully explained (3)

Similar/Related Questions

In the figure, a real inverted image I of an object O is formed by a certain lens (not shown); the object-image separation d = 40.0 cm, measured along the central axis of the lens. The image is exactly half the size of the object. (a) What kind of lens must be used to produce this image? (b) How far from the object must the lens be placed? (c) What is the focal length of the lens?
Solution
0 0

In the figure, a real inverted image I of an object O is formed by a certain lens (not shown); the object-image separation is d = 58.2 cm, measured along the central axis of the lens. The image is just 1/4 the size of the object. (a) What kind of lens must be used to produce this image? (b) How far from the object must the lens be placed? (c) What is the focal length of the lens? (a) (b) Number Units (c) Number Units
Solution
0 0

This problem deals with several aspects of an oscilloscope. You have an 14700 volt supply (Vsupply) for accelerating electrons to a speed adequate to make the front phosphor-coated screen glow when the electrons hit it. Once the electron has emerged from the accelerating region, it coasts through a vacuum at nearly constant speed. You can apply a potential difference of plus or minus 35 volts (Vplates) across the deflection plates to steer the electron beam up or down on the screen to paint a display (other deflection plates not shown in the diagram are used to steer the beam horizontally). Each of the two deflection plates is a thin metal plate of length L = 8 cm and width (into the diagram) 4 cm. The distance between the deflection plates is s = 5 mm. The distance from the deflection plates to the screen is d = 34 cm. When there is a 35 volt potential difference between the deflection plates, what is the deflection y of the electron beam where it hits the screen? An approximate treatment is fine, but state your assumptions. As is usually the case, it pays to carry out all of your calculations algebraically and only evaluate the final algebraic result numerically. Note the exaggerated vertical scale: the deflection is actually small compared to the distance to the screen. When leaving the accelerating plates, the electron's speed is close to the speed of light. Therefore, you must use K = (γ − 1)mc2 where γ = 1 /1 − (v/c)2 and c is the speed of light, 3×108 m/s. m
Solution
0 0