An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Answers
Answer:
Postition of image = At a distance 16.66 cm away from lens on right side.
Size of image = 3.3 cm on right side of lens
Nature of image = inverted and real.
Explanation:
Given,
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm.
To find: The position(v), size and the nature of the image formed.
Solution:
From sign convention,
Distance of image(u) = - 25 cm
Height of object (H_o) = + 5cm
Focal length of the lens ( f) = + 10 cm
Applying Len's formula,
1/v - 1/u = 1/f
1/v - 1/(-25) = 1/10
1/v = 1/10 - 1/25
1/v = (5 - 2)/50
1/v = 3/50
or, v= 50/3 = 16.66 cm
From Magnification formula,
m = - v / u = -16.66/25 = -0.66
Also, magnification can be given as:
m = height of image / height of object
or, m = H_i / H_o
or, -0.66 = H_i / 5
or, H_i = -0.66 * 5 = -3.3 cm
Since, Height of image is negative we can say that image formed will be inverted and real. Ray diagram is attached.
Therefore, Postition of image = At a distance 16.66 cm away from lens on right side.
Size of image = 3.3 cm on right side of lens
Nature of image = inverted and real.
An object 5cm high is placed at distance of 60cm in front of concave mirror of focal length 10cm. Find position and size of image by drawing?
Prajitha Aniyeri
Answered 3 years ago
Object distance , u= -60cm
Focal lenght , f =-10cm
Object size = 5 cm
v =?
Mirror formula,
1/v + 1/u = 1/f
1/v = 1/u - 1/f
1/v= 1/-(60) +1/-(10)
1/v = -1+6/60
1/v = -5/60
1/v = -1/12
v = -12
Image position = -12
Size of the image,
Magnification , m= -v/u
m = -(-12)/-60
m = 12/-60
m = 1/-5
m= -0.2
Then ,
m = h'/h
-0.2= h'/5
-0.2 × 5 = h'
h' = -1