An object is 0.5 cm high and is 6.0 cm from a concave lens with a radius of curv

Question

An object is 0.5 cm high and is 6.0 cm from a concave lens with a radius of curvature of 12.0 cm. What is the focal length, image location, magnification and height or image? Green light has a wavelength of about 5.40 times 10^-7 m. What is its frequency? If Bob Bungee bounced up and down 5 times in one minute while hanging from the end of a bungee cord, what was the frequency of his vibration in Hertz? The image of a candle is seen 22 cm from a convex lens when the candle is placed 18 cm from the lens. What is the focal length of the lens? What is the magnification of the lens? If your nose is 5 cm long but its image is 20 cm long in a concave mirror, what is the mirror's magnification? A convex lens has a radius of curvature of 32 cm. What is its focal length? If a candle was placed 28 cm from the lens, what is the image distance? An object is located 10 cm in front of a concave lens that has a focal length of -8.0 cm. What is the distance from the lens to the image?

Explanation / Answer

Given

50
  
Height of the object 0.5 cm ,

object distance do = 6 cm,

inage distance di = ?

focal length of the concave lens is f = R/2 =12.0 /2 = - 6 cm

image location is

   1/di = 1/f - 1/d0

   = - 1/6 - 1/6

   di = -3 cm

magnification is m = -di/do = hi/ho

       = - (-3)/6
       = 0.5 cm

height of the image of the object is hi = m*h0 = 0.5*0.5 = 0.25 cm

51
wavelength of the green light is 5.4*10^-7 m

the relation between the veloity and the wavelength and frequency is

   v = lambda*f
   f = v/lambda

   f = 3*10^8 /( 5.4*10^-7) Hz

   F = 5.56*10^14 Hz
  
52.
bob bounced up and down 5 times in one min so t

the frequency is no of oscillations per sec

   f = 5/60 = 0.0833 Hz
53.

   given convex lens

   of vocal length f = ?

   object distance d0 = 18 cm , image distance di =22 cm
from the formula

   1/f = 1/do + di
  
   = 1/18 + 1/22

   f = 9.9 cm

54.

   magnification of lens is m = -di/do
               m = -22/18

               m = - 1.22

55.

   mirrors magnification i s m = -di /do
               m = - 20/5 = -4
-ve sign is --- virtual image

56.

the relation between the focal length and the radius of curvature is

   f = R/2 = 32/2 = 16 cm

57 . do = 28 cm

   di = ?

   f = 16 cm

   1/di = 1/f - 1/do

   1/di = 1/16 - 1/28

   di = 37.33 cm
image distance is di = 35.33 cm

58
   1/di = 1/f +1/do

       = -1/8 + 1/10

   di = -0.025 cm

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