1)An object of height 4 cm is placed at a distance of 20 cm from a convex lens of focal length 10 cm. Find the nature of the image.
1/f=1/d1+1/do
d1=fdo/do-f
(10)(20)/20-10
d1=20
m=-(20)(4)/20
-4 or 4 away
answer: real, inverted
2)Light passes from air into water at 45.0 d/C to the normal. If the index of refraction of water is 1.33, find the angle of refraction.
sin-1((1.00/1.33)sin45.0
32.1 d/C
The numbers are correct, but I am not familiar with your "d/C" notation. I hope it has nothing to do with degrees C, which is a unit of temperature, not angle. Angle-measuring units are just called "degrees".
To find the nature of the image formed by a convex lens, we can use the lens formula:
1/f = 1/d1 + 1/do
where:
- f is the focal length of the lens
- d1 is the distance of the object from the lens (taken as positive if on the same side as the incident light)
- do is the distance of the object from the lens (taken as positive if on the same side as the incident light)
In this case, the object is placed at a distance of 20 cm from a convex lens with a focal length of 10 cm.
Substituting the given values into the formula:
1/10 = 1/d1 + 1/20
To find the value of d1, we can solve this equation. First, let's eliminate the fractions by finding the common denominator:
20/d1 + 20/20 = 1/10
Now, let's simplify:
20/d1 + 1 = 1/10
Next, let's isolate the variable d1:
20/d1 = 1/10 - 1
Combining the fractions:
20/d1 = 1/10 - 10/10
20/d1 = -9/10
Let's invert both sides of the equation to isolate d1:
d1/20 = -10/9
Now, let's solve for d1 by multiplying both sides by 20:
d1 = (20)(-10/9)
Simplifying:
d1 = -200/9 = -22.22 cm (approximately)
The negative sign indicates that the image is formed on the same side as the incident light (virtual image).
To determine the nature of the image, we can use the magnification formula:
m = -d1/do
where:
- m is the magnification (taken as positive for an upright image and negative for an inverted image)
- d1 is the distance of the image from the lens
- do is the distance of the object from the lens
Substituting the values:
m = -(-22.22 cm) / 20 cm
m = 1.11
Since the magnification is positive, the image is upright. Therefore, the nature of the image is virtual and upright.
Note: The calculations above were done assuming the object is placed on the same side as the incident light. If the object is placed on the other side (opposite to the incident light), the sign conventions for the distances would change, leading to different results.