A convex meniscus lens is made of glass with refractive index, n= 1.5. The radius of curvature of the concave surface is 46.2 cm and the radius of curvature of the convex surface is 22.4 cm.
a)Calculate the focal length of the lens.
b)Determine the position of the image of an object which is placed 2.0 m in front of the lens.
n=1.5
r1=22.4cm
r2 = 46cm
a)
(1/f) = (n-1) ( (1/r1) -(1/r2))
(1/f)= (1.5-1) ((1/22.4) - (1/46) )
f= 87.32 cm
focal length of lens is 87.32 cm
a) Well, calculating the focal length of a convex lens is as exciting as finding a needle in a haystack, but lucky for you, I'm here to make it less tedious. Hold onto your seat, because here's the formula:
1/f = (n - 1) * (1/R1 - 1/R2)
Where f is the focal length, n is the refractive index, R1 is the radius of curvature of the concave surface, and R2 is the radius of curvature of the convex surface. Now, plug in the values and let the math games begin!
1/f = (1.5 - 1) * (1/46.2 - 1/22.4)
Now, solve the equation and you'll find the focal length of the lens. Good luck, math wizard!
b) Ah, the position of the image question. Let's see if I can perform some magic and give you the answer. When an object is placed in front of a lens, the image forms on the other side of the lens. In this case, the object is 2.0 m in front of the lens, so the image will form...drumroll, please...at a distance equal to the focal length from the lens!
So, if you calculated the focal length in part a), all you have to do is add that value to 2.0 m, and voila! You've got the position of the image. Hang on tight, the answer is coming your way!
To calculate the focal length of the convex meniscus lens, we can use the lens maker's formula:
1/f = (n - 1) * ((1 / R1) - (1 / R2))
where:
- f is the focal length of the lens
- n is the refractive index of the lens material (in this case, n = 1.5)
- R1 is the radius of curvature of the concave surface
- R2 is the radius of curvature of the convex surface
a) Calculate the focal length of the lens:
Using the given values, we have:
1/f = (1.5 - 1) * ((1 / -46.2) - (1 / 22.4))
Simplifying further:
1/f = (0.5) * ((-22.4 + 46.2) / (-46.2 * 22.4))
1/f = (0.5) * (23.8 / (-46.2 * 22.4))
Calculating:
1/f = (0.5) * (-0.00952)
1/f = -0.00476
Taking the reciprocal of both sides:
f = -1 / (-0.00476)
f = 209.24 cm
Therefore, the focal length of the lens is 209.24 cm.
b) To determine the position of the image of an object placed 2.0 m in front of the lens, we can use the lens formula:
1/f = 1/v - 1/u
where:
- f is the focal length of the lens (which we calculated to be 209.24 cm)
- v is the distance of the image from the lens (unknown)
- u is the distance of the object from the lens (2.0 m)
Plugging in the values, we get:
1/209.24 = 1/v - 1/200
To solve for v, we rearrange the equation:
1/v = 1/209.24 + 1/200
1/v = (200 + 209.24) / (209.24 * 200)
1/v = 409.24 / 41850.9
Calculating:
1/v = 0.00979
Taking the reciprocal of both sides:
v = 1 / 0.00979
v = 102.1 cm
Therefore, the position of the image is 102.1 cm from the lens.
To calculate the focal length of the lens, we can use the lens maker's formula:
\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where:
- \(f\) is the focal length
- \(n\) is the refractive index of the lens material (in this case, glass with \(n = 1.5\))
- \(R_1\) is the radius of curvature of the concave surface
- \(R_2\) is the radius of curvature of the convex surface
a) Given:
\(n = 1.5\)
\(R_1 = 46.2 \, \text{cm} \)
\(R_2 = 22.4 \, \text{cm} \)
We can substitute these values into the lens maker's formula to find \(f\):
\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{46.2} - \frac{1}{22.4} \right) \]
Simplifying the expression:
\[ \frac{1}{f} = 0.5 \left( \frac{22.4 - 46.2}{46.2 \times 22.4} \right) \]
\[ \frac{1}{f} = 0.5 \left( \frac{-23.8}{1034.08} \right) \]
\[ \frac{1}{f} = -0.0116 \]
\[ f = \frac{1}{-0.0116} \]
\[ f \approx -86.21 \, \text{cm} \]
The focal length of the lens is approximately -86.21 cm.
b) To determine the position of the image of an object placed 2.0 m in front of the lens, we can use the lens equation:
\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]
where:
- \( f \) is the focal length
- \( d_o \) is the object distance (in this case, 2.0 m)
- \( d_i \) is the image distance (the distance from the lens to the image)
We can rearrange the formula to solve for \( d_i \):
\[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \]
Plugging in the values:
\[ \frac{1}{d_i} = \frac{1}{-86.21 \, \text{cm}} - \frac{1}{200 \, \text{cm}} \]
Simplifying:
\[ \frac{1}{d_i} = \frac{-200 + 86.21}{(-86.21) \times 200} \]
\[ \frac{1}{d_i} = \frac{-113.79}{-17242} \]
\[ \frac{1}{d_i} = 0.0066 \]
\[ d_i = \frac{1}{0.0066} \]
\[ d_i \approx 151.52 \, \text{cm} \]
Therefore, the image of the object placed 2.0 m in front of the lens appears at a distance of approximately 151.52 cm from the lens.