Question

Prelab Name Class Section The image of an object through a lens can be found as the intersection of two light rays starting from the tip of the object. Any two rays can do. Two convenient ones are: 1) the ray passing through the center of the lens, as it does not get deflected; 2) the ray perpendicular to the axis of the lens, as it gets deflected through the focus of the lens. The attached drawing shows an object as an arow on the left side and a lens as a double arrow at the center. The lens focal points are also shown. Using your ruler and referring to the scale indicated on the drawing, find: a) The distance (in mm) of the image from the lens b) The size (in mm) of the image. c) Is the image right side up or inverted? d) Is the image real or virtual? e) Is the distance of the image from the lens (as measured on the graph paper) in accord with eq. (1) from the handout? Calculate the % error. f The magnification (ratio of the image size to the the size of the object) equals q/p. Does the size of your image (as measured on the graph paper) agree with the above formula? Calculate the % error.

Answer 1

Done the ray diagram in plain paper but the distances are marked properly so reproducing won't be a problem.

Given Focal length 2 cm. Object distance =5 cm, Object size = 2 cm.

a) Image distance =33 mm.(See in diagram)

3.3m distance

b) Size of image =13 mm.(See in diagram)

c) The image is inverted.

d) The image is real.(As formed by actual joining of the light rays and not virtially coming from a point.)

e)

$\frac{1}{f} =\frac{1}{d_{o}} + \frac{1}{d_{i}}$

From the above equation the image distance is 33.33 cm

Error =(33.33 -33)/33.33=0.01 , Percentage Error =1%

f) Magnification = image size/object size = -13/20 = -0.65

from the formula $M=- \frac{d_{i}}{d_{o}}$ = -2/3

Error = (-0.6667+0.65)/-0.6667 = 0.025, Percentage Error =2.5 %