(GR0177 #11) Lens Question on GR0177
(GR0177 #11) Lens Question on GR0177
ETS Practice Test: GR0177
Question 11
I have tried solving this problem by using the thin lens equation:
1/s + 1/s1 = 1/f
For the first lens, the image is 40cm after the second lens. Now using a value of 10cm [40cm  30cm] as object distance s for the second lens, this yields an image distance of infinity to the left of the second lens.
This unfortunately is not amongst the options...
Could someone please explain how to do this?
Question 11
I have tried solving this problem by using the thin lens equation:
1/s + 1/s1 = 1/f
For the first lens, the image is 40cm after the second lens. Now using a value of 10cm [40cm  30cm] as object distance s for the second lens, this yields an image distance of infinity to the left of the second lens.
This unfortunately is not amongst the options...
Could someone please explain how to do this?
Thanks! I managed to figure it out. I don't think you can use the LensMake equation here. I looked it up and you need the radii of curvature of the lenses to use this formula.
But you can solve the problem by using the thin lens equation. What I was doing wrong was the signs. There is a sign convention that you have to apply when you use the thin lens equation. You have:
1/s + 1/S = 1/f
where s is the object distance, S is the object distance and f is the focal length. According to the sign convention, any distance left of the lens is negative and any distance right to the lens is positive.
Using this convention, you get to the right answer.
Where are you taking your GREs btw? Mine is on tuesday the 5th April.
But you can solve the problem by using the thin lens equation. What I was doing wrong was the signs. There is a sign convention that you have to apply when you use the thin lens equation. You have:
1/s + 1/S = 1/f
where s is the object distance, S is the object distance and f is the focal length. According to the sign convention, any distance left of the lens is negative and any distance right to the lens is positive.
Using this convention, you get to the right answer.
Where are you taking your GREs btw? Mine is on tuesday the 5th April.
GRE Test Date
I take mine on April 2nd, as in tomorrow. *gulp*
If I can recall any questions after I drink myself into oblivion after the test, I'll post them here. Usually I can remember 1 or 2 verbatim..,
Good Luck on yours...
If I can recall any questions after I drink myself into oblivion after the test, I'll post them here. Usually I can remember 1 or 2 verbatim..,
Good Luck on yours...

 Posts: 2
 Joined: Tue Nov 08, 2005 12:32 am
0177 #11 problem
this problem does not make sense with the two suggested methods posted here.
If an object is at twice the focal length, the image will also be at twice the focal length. (on the opposite side of a converging lens).
This means that the final image will be at infinity to the left, as was originally stated...
This is not one of the available answers.
Please tell me I am wrong.
If an object is at twice the focal length, the image will also be at twice the focal length. (on the opposite side of a converging lens).
This means that the final image will be at infinity to the left, as was originally stated...
This is not one of the available answers.
Please tell me I am wrong.
http://grephysics.net will post complete solutions to GR0177 sometimes late tomorrow (or sometimes in the wee hours of wednesday).
For the timebeing, here's the solution I have for this problem:
The lensmaker's formula is $1/d_o+1/d_i=1/f$. For an image on the opposite side of the light, the image distance is taken as positive.
The distance between the object and the first lens is $d_{1o}=40$. $f_1=20$. The lensmaker's formula gives $1/d_{1i}=1/f_11/d_{1o}=1/40$. Thus, the image is 40 cm behind the first lens.
The first image forms the object for the second lens. The distance of the first image to the second lens is $4030=10cm$. Since this image is behind the lens (on the other side of the incident geometric light), the convention in the lensmaker's formula takes this as a \emph{negative} distance. One has, $1/d_{i2}=1/f_2  1/d_{o2}=1/10+1/10=2/10=1/5$. Thus, the second image is 5 cm to the right of the second lens.
For the timebeing, here's the solution I have for this problem:
The lensmaker's formula is $1/d_o+1/d_i=1/f$. For an image on the opposite side of the light, the image distance is taken as positive.
The distance between the object and the first lens is $d_{1o}=40$. $f_1=20$. The lensmaker's formula gives $1/d_{1i}=1/f_11/d_{1o}=1/40$. Thus, the image is 40 cm behind the first lens.
The first image forms the object for the second lens. The distance of the first image to the second lens is $4030=10cm$. Since this image is behind the lens (on the other side of the incident geometric light), the convention in the lensmaker's formula takes this as a \emph{negative} distance. One has, $1/d_{i2}=1/f_2  1/d_{o2}=1/10+1/10=2/10=1/5$. Thus, the second image is 5 cm to the right of the second lens.

 Posts: 2
 Joined: Tue Nov 08, 2005 12:32 am
Where should I start?
Crumbles wrote:
Hello,
I am from NEPAL and thinking to take subject Test. But, I have to start from scartch. Which approcah and how much study is requied for medium score. Also, which book is best and is there any materials easily available expect ETS booklet? I have never dreamt of TOP school. Just medium school is OK for me. My TOEFL score is 250 and GRE general 1140 ( 390 verbal and quant. 750). How much score on subject test score is required?
Could someone please explain how to do this?