Question

Click Submit to complete this assessment. Question 21 of 21 Question 21 4 points Save Answer An object of height 2 cm is 10 cm in front of a diverging with focal length 1f1 = 10 cm The MAGNITUDE OF the image distance in cm is INSTRUCTIONS: PLEASE TYPE A POSITIVE (MAGNITUDE) NUMERICAL ANSWER, NO UNIT REQUIRED. The MAGNITUDE OF the image height in cm 1: INSTRUCTIONS: PLEASE TYPE A POSITIVE (MAGNITUDE) NUMERICAL ANSWER, NO UNIT REQUIRED The image is: (IN FRONT OF BEHIND) lens. INSTRUCTIONS: CHOOSE ONE AND COPY AND PASTE The image is: (ENLARGEDOREDUCED). INSTRUCTIONS: CHOOSE ONE AND COPY AND PASTE The image is: (UPRIGHT/INVERTED). INSTRUCTIONS: CHOOSE ONE AND COPY AND PASTE The image is: (REAL/VIRTUAL), INSTRUCTIONS. CHOOSE ONE AND COPY AND PASTE This is an example of (FARSIGHTED GLASSES/NEARSIGHTED GLASSES / PROJECTOR LENS). INSTRUCTIONS: CHOOSE ONE AND COPY AND PASTE

Answer 1

Let u be the object distance

v be the image distance

f be the focal length

h be the height of the object

h' be the height of the image

Here,

u = 10cm

h = 2cm

f = 10 cm

Using

-1/f = 1/u + 1/v

-1/10 = 1/10 + 1/v

1/v = 2/10

v = -5 cm

therefore

image distance is 5 cm

image height is , h' = h x (-v/u) = 2 x (5/10) = 1 cm

image is in front of the lens

image is reduced

image is upright

image is virtual

farsight glass

I hope this will help you. Have a good day. Thanks.