Image formed by lens-water-mirror system

[ConorDMK]

Homework Statement

A symmetric, double-convex, thin lens made of glass with n = 1.52 has a focal length in air of 40 cm. The lens is sealed into the opening in the left-hand of a tank filled with water (n = 4/3). At the right-hand end of the tank is a plane mirror 90 cm from the lens. Find the position and lateral magnification of the image formed by the lens-water-mirror system of a small object outside the tank on the lens axis and 70 cm to the left of the lens.

Variables:
f=focal length=40cm,
n_a=refractive index of air=1,
n_g=refractive index of glass=1.52,
n_w=refractive index of water=4/3,
x=distance from lens to mirror=90cm,
s₁=object distance=70cm

Homework Equations

1/f=(n_g-n_a)(1/R₁-1/R₂)

n_a/s₁+n_g/s'₁=(n_g-n_a)/R₁

n_g/s₂+n_w/s'₂=(n_w-n_g)/R₂

The Attempt at a Solution

As the lens is symmetrical, R₁=-R₂
⇒ 1/f=(n_g-n_a)(1/R₁-1/R₂)=(n_g-n_a)(1/R₁+1/R₁)
⇒ 1/f=(n_g-n_a)(2/R₁)
⇒ R₁=2f(n_g-n_a)

Using -s'₁=s₂
n_a/s₁+n_g/s'₁=(n_g-n_a)/R₁
⇒s'₁=n_g/(((n_g-n_a)/R₁)-(n_a/s₁))

s₂=n_g/(((n_w-n_g)/(-R₁))-(n_g/s'₂))

⇒-n_g/(((n_g-n_a)/R₁)-(n_a/s₁))=n_g/(((n_w-n_g)/(-R₁))-(n_g/s'₂))

⇒((n_g-n_a)/R₁)-(n_a/s₁)=((n_w-n_g)/R₁)+(n_g/s'₂)

⇒s'₂=n_w/(((n_g-n_a)/R₁)-(n_a/s₁)-((n_w-n_g)/R₁))

And s₃=x-s'₂

But this is where my friends and I get bogged down in where everything is supposed to be.

 

[TSny]

I suggest that you plug in numbers as you go so that you work out each step numerically. I know this goes against the general rule of working all the way to the end symbolically and then substituting numbers, but the algebra in this case is obviously getting out of hand. By plugging the numbers in as you go, you can also see if your answers are making sense as you go. This problem is a field of land mines in regard to signs.

 

[ConorDMK]

I suggest that you plug in numbers as you go so that you work out each step numerically. I know this goes against the general rule of working all the way to the end symbolically and then substituting numbers, but the algebra in this case is obviously getting out of hand. By plugging the numbers in as you go, you can also see if your answers are making sense as you go. This problem is a field of land mines in regard to signs.

That's the line of thinking that we decided to go down in the end. And having done that, I decided to do the full algebraic expression for the final image and final magnification anyway, just to annoy the people marking it. I feel like Dr Frankenstein.

 

[TSny]

That's the line of thinking that we decided to go down in the end. And having done that, I decided to do the full algebraic expression for the final image and final magnification anyway, just to annoy the people marking it. I feel like Dr Frankenstein.