Question

A concave lens of focal length 15cm forms an image 10cm from the lens. How far is the object placed from the lens? Draw the ray diagram. THE BEST ANSWER WILL BE MARKED THE BRAINLIST

Answer 1

Answer:

Given,

focal length, f = -15 cm

Distance of image, v = -10 cm

Distance of object, u =?

We know that,

1/v - 1/u = 1/f

So,

1/-10 - 1/u = 1/-15

So, -1/10 + 1/15 = 1/u

(-3 + 2)/30 = 1/u

-1/30 = 1/u

Therefore u = -30 cm

Thus object is placed 30 cm away from the concave lens.

Negative sign shows that object is at 30cm in front of the lens.

Explanation:

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Answer 2

Answer:

Given, a concave lens.

Focal length, f = - 15 cm [f is - ve for a concave lens]

Image distance, v = - 10 cm [image formed is virtual i.e., on same side as the object, so v is - ve]

Now, using the lens formula,

1f =1v-1u

1u =1v-1f = 1-10-1-15 = -3+230 = -130

Therefore, object distance, u = -30 cm

Ray diagram:

Inorder to make the diagram, lets use a scale where 5cm = 1cm.

So, as per the new scale,

Focal length, f = -3 cm

Image distance, v = -2 cm

Steps to draw the ray diagram is mentioned below as follows:

(i) Draw a horizontal line which is called the principal axis.

(ii) Now, draw a convex lens keeping principal centre (C) on the principal axis.

(iii) Mark points F (focal length) and B (image distane) on the left side of lens at a distance of 3 cm and 2 cm respectively.

(iv) Draw a dotted line passing through F to any point on the top of the lens, say D.

(v) So, we can draw a line AD parallel to principal axis because any ray of light passing through the focal length of the lens after refraction, passes parallel to the principal axis.

(vi) Draw a line A'B', perpendicular to principal axis from B' representing the height of the image.

(vii) Draw a line CA' backwards, so that it meets the line from D at A.

(viii) Now, draw a line AB, perpendicular to the principal axis at B from point A in the downward direction.

(ix) AB is the position of object. On measuring distance BC, it will be found to be equal to 6 cm.

Thus, the object is placed at a distance of 6 cm × 5 = 30 cm from the lens (as per the original scale).