An object 2.00cm high is placed 40cm to the left of a converging lens having a focal length of 30cm.A diverging lens having a focal length of -20cm is placed 110 cm to the right of the converging lens. Determine the final position and magnification of the final image.
Well, well, well, we've got ourselves a lens party going on here! Let's get this optical extravaganza sorted out.
First, we need to tackle the converging lens. We'll use the lens equation 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance.
Given that the object distance u is -40cm (because it's to the left of the lens) and the focal length f is 30cm, we can solve for v:
1/30 = 1/v - 1/(-40)
Now, let's put our equation-solving caps on and crunch those numbers! And after some calculation, we get v = 120 cm.
So, the image formed by the converging lens is located 120 cm to the right of that lens.
Now, moving on to the diverging lens. Since we're dealing with two lenses, we need to find the object distance (u') for the diverging lens. The formula to use here is 1/v' = 1/f' - 1/u', where f' is the focal length of the diverging lens and v' is the image distance created by the converging lens.
Knowing that f' is -20 cm (because it's a diverging lens) and v' is 120cm, let's plug those values in:
1/(-20) = 1/u' - 1/120
Hold the laughter, please! After performing the calculations, we get 1/u' = 1/120 - 1/(-20), or 1/u' = 1/30 + 1/120. Simplifying that further, we get 1/u' = 4/120, or 1/u' = 1/30.
Now, flip that equation upside down (just like my sense of humor) and we find that u' = 30 cm.
So, the object distance for the diverging lens is 30 cm.
Finally, we can determine the image distance (v') and magnification (M) for the diverging lens. Since we're dealing with a diverging lens, we'll use the same lens equation as before.
Given that the object distance u' is 30 cm and the focal length f' is -20 cm, we can solve for v':
1/(-20) = 1/v' - 1/30
Now, it's time again for some equation-solving magic! After calculating the values, we find that v' = -60 cm.
Ah, a negative sign? How rebellious! This means the image formed by the diverging lens is virtual and located 60 cm to the left of the lens.
Finally, let's find the magnification (M). The magnification is given by M = v'/u'. Plugging in the values, we get:
M = -60/30 = -2
Here you have it! The final position of the image is -60 cm (to the left of the diverging lens) and the magnification is -2. Happy optical acrobatics!