A lighted candle is placed 38 cm in front of a diverging lens. The light passes through the diverging lens and on to a converging lens of focal length 10 cm that is 5 cm from the diverging lens. The final image is real, inverted, and 19 cm beyond the converging lens. Find the focal length of the diverging lens.
if a candle is standing in front of a diverging lens , we can see its image on a screen clearly
To find the focal length of the diverging lens, we can use the lens formula:
1/f = 1/v - 1/u
where f is the focal length, v is the image distance, and u is the object distance.
Given:
Object distance (u) = -38 cm (negative because it is on the opposite side of the lens)
Image distance (v) = 19 cm
Substituting these values into the formula:
1/f = 1/19 - 1/(-38)
Simplifying:
1/f = 1/19 + 1/38
1/f = (2 + 1)/38
1/f = 3/38
To isolate f, we take the reciprocal of both sides:
f = 38/3
So, the focal length of the diverging lens is 38/3 cm.
To find the focal length of the diverging lens, we can use the lens formula:
1/f = 1/v - 1/u
where f is the focal length of the lens, v is the image distance, and u is the object distance.
Let's break down the given information:
Object distance of the diverging lens (u1) = -38 cm (negative because it is in front of the lens)
Image distance formed by the diverging lens (v1) = ?
Focal length of the converging lens (f2) = 10 cm
Distance between the diverging lens and the converging lens (d) = 5 cm
Final image distance (v2) = 19 cm
Using the lens formula for the diverging lens:
1/f1 = 1/v1 - 1/u1
As the final image is formed beyond the converging lens, we consider the image formed by the diverging lens (v1) as the object for the converging lens.
Object distance for the converging lens (u2) = v1
Using the lens formula for the converging lens:
1/f2 = 1/v2 - 1/u2
Now, let's calculate the values step by step:
1. Calculate the image distance formed by the diverging lens (v1):
1/f1 = 1/v1 - 1/u1
1/f1 = 1/v1 - 1/(-38 cm)
1/f1 = 1/v1 + (1/38) cm⁻¹
2. Calculate the object distance for the converging lens (u2), considering v1 as the object distance:
u2 = v1
3. Calculate the image distance formed by the converging lens (v2):
1/f2 = 1/v2 - 1/u2
1/10 = 1/19 - 1/u2
Now we have two equations:
Equation 1: 1/f1 = 1/v1 + (1/38) cm⁻¹
Equation 2: 1/10 = 1/19 - 1/u2
We can use these two equations to solve for v1 and u2 simultaneously.
4. Rearrange Equation 2 to solve for u2:
1/10 + 1/u2 = 1/19
1/u2 = 1/19 - 1/10
1/u2 = (10 - 19)/(10 * 19)
1/u2 = -9/190
u2 = -190/9 cm
5. Substitute the value of u2 in Equation 1 and solve for v1:
1/f1 = 1/v1 + (1/38) cm⁻¹
1/f1 = 1/v1 + (1/38) cm⁻¹
1/f1 = 1/v1 - (1/38) cm⁻¹
1/v1 = 1/f1 + (1/38) cm⁻¹
v1 = 1/(1/f1 + (1/38)) cm
6. Finally, substitute the calculated value of v1 in Equation 2 to solve for f1:
1/10 = 1/19 - 1/u2
1/10 = 1/19 - (9/190) cm⁻¹
1/10 = (190 - 9)/(19 * 190) cm⁻¹
1/10 = 181/3610 cm⁻¹
f1 = 3610/181 cm
f1 = 20 cm
Therefore, the focal length of the diverging lens is 20 cm.