A convex lens forms an image of an object on screen. The height of the image is 9 cm. The lens is now displaced until an image is again obtained on the screen. The height of this image is 4 cm. The distance between the object and the screen is 90 cm.a)The distance between the two positions of the lens is 30 cm.b)The distance of the object from the lens is its first position is 36cm.c)The height of the object is 6cm.d)The focal length of the lens is 21.6 cm.The focal length of the lens is 21.6 cm.Correct answer is option 'B,C,D'. Can you explain this answer?

Question

Answer

B) The distance of the object from the lens in its first position is 36 cm.
To find the distance of the object from the lens in its first position, we can make use of the lens formula:

1/f = 1/v - 1/u

Where f is the focal length of the lens, v is the image distance, and u is the object distance.

We are given the height of the image formed on the screen when the lens is in its first position as 9 cm. Since the height of the image is positive, the image is upright. Therefore, the image distance (v) is positive.

We are also given the distance between the object and the screen as 90 cm. Since the height of the image formed on the screen when the lens is displaced is 4 cm, the image distance (v) is still positive.

Using the lens formula, we can write:

1/f = 1/v - 1/u

Let's assume that the distance of the object from the lens in its first position is u1, and the distance of the lens from the screen is v1.

1/f = 1/v1 - 1/u1 ...(1)

We can also assume that the distance of the object from the lens in its second position is u2, and the distance of the lens from the screen is v2.

1/f = 1/v2 - 1/u2 ...(2)

We know that the image distance (v1) in the first position is equal to the object distance (u2) in the second position. Similarly, the image distance (v2) in the second position is equal to the object distance (u1) in the first position.

From the given information, we can deduce the following:

v1 = 90 cm
v2 = 90 cm
h1 = 9 cm
h2 = 4 cm

c) The height of the object is 6 cm.
To find the height of the object, we can make use of the magnification formula:

m = -v/u

Where m is the magnification, v is the image distance, and u is the object distance.

In the first position:

m1 = -v1/u1 ...(3)

In the second position:

m2 = -v2/u2 ...(4)

We know that the magnification is equal to the ratio of the height of the image to the height of the object:

m1 = h1/u1 ...(5)
m2 = h2/u2 ...(6)

From equations (5) and (6), we can write:

h1/u1 = h2/u2

h1/u1 = 9/u1 = 4/u2 ...(7)

From equation (7), we can solve for the height of the object (h1):

9/u1 = 4/u2

9/u1 = 4/(90-u1)

Solving this equation, we find:

u1 = 36 cm

Therefore, the distance of the object from the lens in its first position is 36 cm.

d) The focal length of the lens is 21.6 cm.
Now that we have the object distance (u1) in the first position and the image distance (v1) in the first position, we can use the lens formula to find the focal length (f).

1/f =

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