A 0.54 cm high object is placed 8.5 cm in front of a diverging lenswhose focal l
ID: 1756443 • Letter: A
Question
A 0.54 cm high object is placed 8.5 cm in front of a diverging lenswhose focal length is -7.5 cm. What is the height of theimage?
I use m=-d_i/d_o, where m is magnification. To find m use the lensequation 1/f=1/d_i+1/d_o and solve it for d_i. That will giveme the m and then I just multiply the m*h_o to get theheight image distance. idk if if my algebra waswrong when i solve for d_i but i am getting the problemwrong. i used d_i = 1/(1/f -1/d_o) Can someone plz help mewith this problem?
A 0.54 cm high object is placed 8.5 cm in front of a diverging lenswhose focal length is -7.5 cm. What is the height of theimage?
I use m=-d_i/d_o, where m is magnification. To find m use the lensequation 1/f=1/d_i+1/d_o and solve it for d_i. That will giveme the m and then I just multiply the m*h_o to get theheight image distance. idk if if my algebra waswrong when i solve for d_i but i am getting the problemwrong. i used d_i = 1/(1/f -1/d_o) Can someone plz help mewith this problem?
I use m=-d_i/d_o, where m is magnification. To find m use the lensequation 1/f=1/d_i+1/d_o and solve it for d_i. That will giveme the m and then I just multiply the m*h_o to get theheight image distance. idk if if my algebra waswrong when i solve for d_i but i am getting the problemwrong. i used d_i = 1/(1/f -1/d_o) Can someone plz help mewith this problem?
Explanation / Answer
Given : Object distance (u) = 8.5 cm Focal length (f) = - 7.5 cm From Lens maker's formula we have : 1/ f = 1/ u + 1 / v or 1 / v = 1 / f -1 /u = - 1 / 7.5 - 1 / 8.5 v = -3.98 cm We know that : Magnification (M) = - v /u = 3.98 / 8.5 = 0.468 But M = - hi / ho or hi = M * ho = 0.468 * 0.54 cm = -0.253 cm Hope this helps u!Related Questions
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