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hammer [34]
2 years ago
12

Hi can someone pls answer the question number 2 pls i srsly need it. thank u :) ​

Physics
2 answers:
Anvisha [2.4K]2 years ago
5 0

<em>The distance between the image and its leans is 19.04 cm...</em>

  • <u>Refer</u><u> </u><u>to</u><u> </u><u>the</u><u> </u><u>attachment</u><u> </u><u>for</u><u> </u><u>detailed</u><u> </u><u>solution</u><u>!</u><u>!</u><u>~</u>

ad-work [718]2 years ago
3 0

Answer:1,0.95cm

2,21.82cms

3, 2.73

Explanation:

1,Its equation is following: 1o + 1i = 1f

1i = 1f - 1o

1i = 1−11 - 125

i = -7.64 cm - the image distance from the lens. The negative sign means that the image virtuali

h = 3.1* 7.6425 = 0.95 cm - the diameter of the coin’s imageimage

2,Use the lens equation to determine the image distance: 1/25 = 1/80 + 1/Di →

Di = 36.36 cm.

The image is real and on the opposite side of the lens as compared to the object.

The distance between object and image then is: 80 + 36.36 = 116.36 cm.

3,1/V+1/U =1/F ; 1/V =1/F-1/U ;- 1/15–1/48 =16-5/ 240 = 11/240 ; V =240/11

=21.82 CMS

IMAGE MAGNIFICATION V /U =21.82/48= 0.0.4546INVERTED

6 * 0.4546=2.7276 ≈2.73

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