# Effect of two lenses after single diffraction grating

I am having problems undestanding the effect of two lenses after a diffraction grating of slits whose separation is unknown.

The setup is like this: 3cm away from the slits there's a convergent lens of unknown focal length (first question of the problem), 3cm away from that lens there's a divergent lens of focal length -3cm, after the 2nd lens there's the screen on which the diffraction pattern appears, 6cm to the right of the lens 2.

The last piece of data I'm given is that the linear distance on the screen between maxima of orders 1 and 2 is 2 cm.

The first question was to determine the focal length of lens 1. I've computed the focal length given that for lens 2 to form an image 6cm to the right, the object whose image is forming must be 2cm to its right. So the 1st lens must create the image of the slit (3cm to its left) 5cm to its right, because the separation of lens 1 and 2 is 3cm.

Now, the 2nd question is to determine the separation of the slits. I've computed that said separation must be:

$$d=s\lambda/\Delta x$$

$s$ being the distance from lens 2 to the screen and $d$ the separation of slits and $\Delta x$ the separation between maxima of order 1 and 2.

The problem is I'm not sure the actual separation is $d$ because I think lens 2 must have "magnified" the image created by lens 1, the magnification being given by:

$$\beta^\prime= s^\prime/s$$

where $s^\prime$ is the distance between the image and lens 2, and $s$ the distance between the object (image created by lens 1 and lens 2. In this case $\beta^\prime = 6/2 = 3$.

Is this correct or am I overcomplicating things?

• "3cm to away from that lens there's a divergent lens" (para 2) but "the separation of lens 1 and 2 is 5cm" (para 4). I'm puzzled. – Philip Wood Jan 8 '18 at 23:51
• Sorry, that was definitely a typo. I’ve corrected it already. – Mikel García Jan 8 '18 at 23:57
• Paragraph 3 - The linear distance between orders 1 and 2 is missing? – Farcher Jan 9 '18 at 7:52

## 1 Answer

Consider what would you would see on a screen which was $5\,\rm cm$ away from the converging lens alone.
You would get a series of maxima and these could be thought of as a real image formed on the screen.

What does introducing the diverging lens do?

The real image which was formed on the screen by the converging lens now becomes a virtual object for the diverging lens and after light passes through the diverging lens a real image is formed at the new position of the screen.
That real image is larger than the virtual object and so the separation between the orders is indeed increased as compared with the separation between the orders when using the converging lens alone.

• So the computed distance between slits, $d$, would be larger than the actual one? – Mikel García Jan 9 '18 at 14:02
• @MikelGarcía Working backwards whatever is the order separation with the diverging lens you can find the order separation if only the converging lens was present. Then it is a standard calculation to find the slit separation but you will need to know the wavelength of the light. – Farcher Jan 9 '18 at 14:08
• Yes, I forgot about the wavelenght $\lambda=632.8 nm$. So because the magnification is $\beta^\prime=3$ the actual distance between maxima on the image formed only by the converging lens is $2/3$, and so I would get the distance between slits just by using $d=f'\lambda/\Delta x$ where $\Delta x=2/3$ and $f'$ is the focal lenght of lens 1. Is that correct? – Mikel García Jan 9 '18 at 14:20