The image of an object formed by a lens is of magnification – 1. If the distance between the object and its image is 60 cm, what is the focal length of the lens ? If the object is moved 20 cm towards the lens, where would the image be formed ? State reason and also draw a ray diagram in support of your answer.

Part 1: Finding the Focal Length of the Lens

Given:

• Magnification, $m = -1$
• Since the magnification is negative, the image is real and inverted. This indicates that the lens is a convex lens (as a concave lens only forms virtual and erect images with positive magnification).
• For a lens, magnification is given by:

$$m = \frac{v}{u} = -1 \implies v = -u$$

• The distance between the object and its real image is $60\text{ cm}$. For a real image formed by a convex lens, the object and the image lie on opposite sides of the lens.

$$\text{Distance} = |u| + v = 60\text{ cm}$$

Since $v = -u$, we have $|u| = v$. Therefore:

$$v + v = 60\text{ cm} \implies 2v = 60\text{ cm} \implies v = 30\text{ cm}$$

Thus, the object distance is $u = -30\text{ cm}$.

Using the lens formula:

\frac{1}{f} = rac{1}{v} - \frac{1}{u}

$$\frac{1}{f} = \frac{1}{30} - \frac{1}{-30} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$$

$$f = +15\text{ cm}$$

Thus, the focal length of the convex lens is $+15\text{ cm}$.

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Part 2: Object Moved $20\text{ cm}$ Towards the Lens

If the object is moved $20\text{ cm}$ towards the lens:

• New object distance, $u' = -30\text{ cm} + 20\text{ cm} = -10\text{ cm}$.
• Focal length, $f = +15\text{ cm}$.

Using the lens formula to find the new image position ($v'$):

$$\frac{1}{f} = \frac{1}{v'} - \frac{1}{u'}$$

$$\frac{1}{15} = \frac{1}{v'} - \frac{1}{-10}$$

$$\frac{1}{v'} = \frac{1}{15} - \frac{1}{10} = \frac{2 - 3}{30} = -\frac{1}{30}$$

$$v' = -30\text{ cm}$$

Reason: Since the new object distance ($10\text{ cm}$) is less than the focal length ($15\text{ cm}$), the object lies between the optical centre ($O$) and the principal focus ($F_1$) of the convex lens. In this position, a convex lens forms a virtual, erect, and magnified image on the same side of the lens as the object.

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Part 3: Ray Diagram

Below is the ray diagram showing the formation of a virtual, erect, and magnified image when the object is placed between $F_1$ and the optical centre $O$:

``diagram

Virtual Image (erect, magnified)

^

| Convex Lens

Object
^

-------------------|--------|--------|--------F1-------O--------F2-----

-30cm -10cm ||

<------ u ------>
<--------------- v' -------------->

``