- #1

Florian Geyer

- 94

- 16

- Homework Statement
- I am doing a problem on geometrical optics in which I need to find the image and object distances a curved refracting surface, the problem contains many parts, but I have problem only with the last two parts, which asked whether it is possible for the image distance to be greater than the image distance.

- Relevant Equations
- Lensmaker's equation for a single surface: ##\frac{n_1}{p} + \frac{n_2}{q} = \frac{n_2 - n_1}{R}##

Magnification formula: ##M = -\frac{n_1 q}{n_2 p}##

Hello esteemed members,

I hope this message will find you well,

I have encountered this problem in Serway and Jewitt 9e.

My problem is only with parts (d) and (e).

I tried to solve the problem in two ways, one is based on intuition, and the second is based on the equations, but the two method did not come hand in hand, which I think clearly means that I made a mistake.

Here is my attempt.

d) Yes this base of the fact that when light goes perpendicular form of medium of Greater n1 to another smaller n2, the image seems to be nearer to the surface for an observer on the second medium, and vice versa.

In this problem light travels between three mediums first between water and plastic and then from plastic to air.

From water to plastic ##n_1 < n_2## The image appears to be further from the wall of the aquarium (from the observer) than it really is.

From plastic to air ##n_1 > n_2## The image appears to be near to The Observer then it really is.

If the plastic wall is very thick (long from front to back), then the effect of this part is greater than the effect of water, does the image appears nearer to the observer.

e) ##\frac{n_1}{p} + \frac{n_2}{q} = \frac{n_2 - n_1}{R} \\

\Rightarrow q = n_2 \left( \frac{n_2 - n_1}{R} - \frac{n_1}{p} \right)^{-1} \\

(n_1 > n_2): (\text{from plastic to air}) ##

##R < 0 \quad (\text{neg}) ##

##n_2 - n_1 < 0 \quad (\text{neg})##

##\frac{n_2 - n_1}{R} > 0 \quad (\text{pos})##

##q = \left( \frac{n_2 - n_1}{R} - \frac{n_1}{p} \right)^{-1}##

(This is the difference between two positive numbers, but as the second term becomes bigger and bigger the quantity ##q## becomes negative)\\

We will study the two extremes when ##R\gg p \,## and when ##R\ll p##,

since all possible values of ##q## will be in between these two extremes.

i) ##p \ll R##:

##\Rightarrow q\approx n_2 (\frac{n_2-n_1}{R})^{-1}=(\frac{n_2 R}{n_2 -n_1})##

##m=\frac{-n_1 q}{n_2 p}##

##\Rightarrow m=- \frac{n_1(n_2 R)/(n_2-n_1)}{n_2 p}##

##\Rightarrow m=-\frac{n_1 R}{p(n_2-n_1)}=0 \quad## (since ##p\ll R##)

ii) ##p \gg R##:

##q \approx n_2 (\frac{-p}{n_1})

m=\frac{-n_1 q}{n_2 p}\Rightarrow m=-\frac{-n_1(-n_2 p/n_1)}{n_2 p}

\Rightarrow m=+1

##

The meaning of the previous values of ##m## is that ##q## can be very small compared to ##p## or equal to it.

The previous was the case from air to plastic.

after this we need to take the other boundary (from water to plastic) into consideration:

if we did the same for the water to plastic case, we will find out that:

##m=0 \quad \text{or} \quad m=1##

In both cased the image distance can either be equal to the body distance or very small in comparison to it.

This result from the math is in contradiction with the one I put first based on my intuition. The solutions manual also does not agree with the result gotten from the math, so please can you help me to understand where did I made mistakes?

I hope this message will find you well,

I have encountered this problem in Serway and Jewitt 9e.

My problem is only with parts (d) and (e).

I tried to solve the problem in two ways, one is based on intuition, and the second is based on the equations, but the two method did not come hand in hand, which I think clearly means that I made a mistake.

Here is my attempt.

d) Yes this base of the fact that when light goes perpendicular form of medium of Greater n1 to another smaller n2, the image seems to be nearer to the surface for an observer on the second medium, and vice versa.

In this problem light travels between three mediums first between water and plastic and then from plastic to air.

From water to plastic ##n_1 < n_2## The image appears to be further from the wall of the aquarium (from the observer) than it really is.

From plastic to air ##n_1 > n_2## The image appears to be near to The Observer then it really is.

If the plastic wall is very thick (long from front to back), then the effect of this part is greater than the effect of water, does the image appears nearer to the observer.

e) ##\frac{n_1}{p} + \frac{n_2}{q} = \frac{n_2 - n_1}{R} \\

\Rightarrow q = n_2 \left( \frac{n_2 - n_1}{R} - \frac{n_1}{p} \right)^{-1} \\

(n_1 > n_2): (\text{from plastic to air}) ##

##R < 0 \quad (\text{neg}) ##

##n_2 - n_1 < 0 \quad (\text{neg})##

##\frac{n_2 - n_1}{R} > 0 \quad (\text{pos})##

##q = \left( \frac{n_2 - n_1}{R} - \frac{n_1}{p} \right)^{-1}##

(This is the difference between two positive numbers, but as the second term becomes bigger and bigger the quantity ##q## becomes negative)\\

We will study the two extremes when ##R\gg p \,## and when ##R\ll p##,

since all possible values of ##q## will be in between these two extremes.

i) ##p \ll R##:

##\Rightarrow q\approx n_2 (\frac{n_2-n_1}{R})^{-1}=(\frac{n_2 R}{n_2 -n_1})##

##m=\frac{-n_1 q}{n_2 p}##

##\Rightarrow m=- \frac{n_1(n_2 R)/(n_2-n_1)}{n_2 p}##

##\Rightarrow m=-\frac{n_1 R}{p(n_2-n_1)}=0 \quad## (since ##p\ll R##)

ii) ##p \gg R##:

##q \approx n_2 (\frac{-p}{n_1})

m=\frac{-n_1 q}{n_2 p}\Rightarrow m=-\frac{-n_1(-n_2 p/n_1)}{n_2 p}

\Rightarrow m=+1

##

The meaning of the previous values of ##m## is that ##q## can be very small compared to ##p## or equal to it.

The previous was the case from air to plastic.

after this we need to take the other boundary (from water to plastic) into consideration:

if we did the same for the water to plastic case, we will find out that:

##m=0 \quad \text{or} \quad m=1##

In both cased the image distance can either be equal to the body distance or very small in comparison to it.

This result from the math is in contradiction with the one I put first based on my intuition. The solutions manual also does not agree with the result gotten from the math, so please can you help me to understand where did I made mistakes?