An object is placed at a distance of 50cm in front of a mirror of focal length 25cm. Find the image distance. If the object is 10cm high what will be the nature and size of the mirror?
Draw a picture
parallel ray from top bounces off and heads through center line at 25 cm from mirror
ray through focus from top comes out parallel below
rays intersect right below original
so:
same distance from mirror
same height but upside down.
check the old answers below by the way.
To find the image distance for an object placed in front of a mirror, we can use the mirror formula:
1/f = 1/u + 1/v
Where:
- f is the focal length of the mirror
- u is the object distance (distance of the object from the mirror)
- v is the image distance (distance of the image from the mirror)
Given:
f = 25 cm
u = 50 cm
Let's substitute these values into the mirror formula:
1/25 = 1/50 + 1/v
To solve for v, we can simplify the equation:
1/v = 1/25 - 1/50
1/v = 2/50 - 1/50
1/v = 1/50
v = 50 cm
The image distance is 50 cm.
Now, let's determine the nature and size of the mirror based on the image distance and the object height.
Since the image distance (v) is positive and greater than the object distance (u), the image formed is a real image.
To find the size of the image, we can use the magnification formula:
m = -v/u
Given:
u = 50 cm
v = 50 cm
Let's substitute these values into the magnification formula:
m = -50/50
m = -1
Since the magnification (m) is negative, it means the image is inverted.
To determine the size of the mirror, we compare the heights of the object (h) and the image (h'):
h'/h = -v/u
Given:
h = 10 cm
u = 50 cm
v = 50 cm
Substituting the values:
h'/10 = -50/50
h' = -10 cm
The size of the image is -10 cm, which means it is 10 cm in height and inverted.
Considering the nature and size of the mirror:
- The mirror is concave because the object is placed closer to it than its focal length, and the image is real and inverted.
- The mirror is of a smaller size than the object (10 cm compared to the object's 10 cm in height).