Calculating distance of image from lens

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  • #1
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Homework Statement




An object is placed on the optic axis and in front of a converging lens of focal length 38 cm such that its image size is 1.5 times the object size. Calculate the distance of the image from the lens. Express your answer in centimeter.

Homework Equations



1/f=1/di + 1/do

M=hi/ho=-di/do



The Attempt at a Solution



Image size is 1.5 times the object size. So 1.5=-di/do
So , do=-di/1.5
So , 1/38=1/di + 1/(-di/1.5) = 1/di - 1.5/di =-0.5/di
Thus di=-0.5*38= -19 .
But the answer is wrong . Please help.
 

Answers and Replies

  • #2
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Len's Formula,
1/f=1/di-1/do
when sign convention is used.

& the formula which u are using

1/f=1/di+1/do
when sign convention is not used.

Magnification,

for real image
m= -hi/ho

for virtual image
m= hi/ho

now u must first clarify that is the image real or virtual.
 
Last edited:
  • #3
ehild
Homework Helper
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1,897

Homework Statement




An object is placed on the optic axis and in front of a converging lens of focal length 38 cm such that its image size is 1.5 times the object size. Calculate the distance of the image from the lens. Express your answer in centimeter.

Homework Equations



1/f=1/di + 1/do

M=hi/ho=-di/do



The Attempt at a Solution



Image size is 1.5 times the object size. So 1.5=-di/do
So , do=-di/1.5
So , 1/38=1/di + 1/(-di/1.5) = 1/di - 1.5/di =-0.5/di
Thus di=-0.5*38= -19 .
But the answer is wrong . Please help.
The problem does not specify if the image is real or virtual. The magnification is defined as M=h(image)/h(object). The real image is inverted, so h(image) is negative. Therefore M=h(image)/h(object) = -1.5 = -di/do, that is, di/do=1.5.

In case of a virtual image, the image is erect, the magnification is positive, so di/do=-1.5, and the image distance is negative. Your result corresponds to that situation, and it is correct, if you explain that you assumed virtual image. The full solution has to include both possibilities: Determine the image distance also for the real image.

ehild
 
  • #4
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The problem does not specify if the image is real or virtual. The magnification is defined as M=h(image)/h(object). The real image is inverted, so h(image) is negative. Therefore M=h(image)/h(object) = -1.5 = -di/do, that is, di/do=1.5.

In case of a virtual image, the image is erect, the magnification is positive, so di/do=-1.5, and the image distance is negative. Your result corresponds to that situation, and it is correct, if you explain that you assumed virtual image. The full solution has to include both possibilities: Determine the image distance also for the real image.

ehild
Awesome ! Taking di/do=1.5 I get the answer as 95 , which is correct.
But how to determine whether the image is real or virtual ?
 
  • #5
ehild
Homework Helper
15,510
1,897
Awesome ! Taking di/do=1.5 I get the answer as 95 , which is correct.
But how to determine whether the image is real or virtual ?
This problem did not specify if the image is real or virtual. If the orientation of the image is given, you can decide. If the problem text was " The image of the object is inverted and its size is 1.5 times the object height" you would know that it was real. If the lens is diverging, you know the image is virtual. If you know that the object is placed closer to the converging lens as its focus, you know that the image is virtual.


ehild
 

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