4) An object is located 15 cm away from the vertex of a bi-convex lens with the focal length of 40 cm. Find out the location of its image, magnification and whether it's virtual or real, and upright or inverted
To determine the location of the image, magnification, and nature (virtual or real) of the image formed by a bi-convex lens, we can use the lens formula and magnification formula.
The lens formula is given by:
1/f = 1/v - 1/u
Where:
f = focal length of the lens
v = image distance (distance of the image from the lens)
u = object distance (distance of the object from the lens)
In this case, the focal length (f) is given as 40 cm, and the object distance (u) is 15 cm. We need to find the image distance (v).
Plugging the values into the lens formula:
1/40 = 1/v - 1/15
Now let's solve this equation to find the value of v.
First, we need to find a common denominator:
1/40 = (15 - v)/(15v)
Next, let's simplify:
1/40 = (15 - v)/(15v)
Multiply both sides of the equation by 15v to eliminate the denominators:
15v/40 = 15 - v
Expand and rearrange the equation:
15v = 40(15 - v)
15v = 600 - 40v
Combine like terms:
15v + 40v = 600
55v = 600
v = 600/55
v ≈ 10.91 cm
So, the image is located at approximately 10.91 cm from the lens.
To find the magnification, we use the magnification formula:
m = -v/u
Where:
m = magnification
v = image distance (already calculated as approximately 10.91 cm)
u = object distance (given as 15 cm)
Substituting the values we have:
m = -(10.91 cm)/15 cm
m ≈ -0.73
The negative sign indicates that the image is inverted.
Lastly, to determine the nature (virtual or real) of the image, we can use the signs of the focal length and the image distance.
Since the focal length of the lens is positive (40 cm) and the image distance (10.91 cm) is positive, the image is real.
In summary:
- The location of the image is approximately 10.91 cm from the lens.
- The magnification is approximately -0.73 (inverted image).
- The image is real.