Find the distance d that places the final image at infinity.

[justagirl]

Anyone know how to solve the following problems? Thanks!

1.) An object is placed 12 cm to the left of a diverging lens of focal length
-6cm. A converging lens of focal length 12 cm is placed a distance of d to the right of the diverging lens. Find the distance d that places the final image at infinity.

2.) Light of wavelength 460 nm falls on a diffraction grating with 5000 slits per cm. What is the required ditance from the slits to a screen if the spacing between the first and second dark fringes is to the 0.7m?

 

[Doc Al]

The first problem is a straightforward application of the lens equation:
$$\frac{1}{f} = \frac{1}{o} + \frac{1}{i}$$

The second problem requires the equation for diffraction:
$$d sin\theta = m \lambda$$

To learn what these equations mean and how to use them, consult your text.

 

[M98Ranger]

To solve the first problem, we can use the thin lens equation: 1/f = 1/di + 1/do, where f is the focal length, di is the image distance, and do is the object distance. Since we want the final image to be at infinity, di = infinity, so the equation becomes 1/f = 0 + 1/do. Plugging in the given values, we get 1/-6 = 1/(12+d). Solving for d, we get d = 18 cm.

For the second problem, we can use the equation for the spacing between fringes in a diffraction grating: d*sinθ = m*λ, where d is the slit spacing, θ is the angle of diffraction, m is the order of the fringe, and λ is the wavelength of light. Since we are given the spacing between the first and second dark fringes, we can set m = 1. Plugging in the given values, we get d*sinθ = 1*460 nm = 0.00046 m. To find the distance to the screen, we can use the small angle approximation: sinθ ≈ θ ≈ tanθ. Therefore, d*θ ≈ d*tanθ = d*(0.00046/0.7) = 0.000657 d. Setting this equal to the given distance of 0.7 m, we get d = 1065.9 cm, or approximately 10.7 m.