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# One half of a convex lens of focal length **10 cm **is covered with black paper. Can such a lens produce an image of a complete object placed at a distance of** 30 cm** from the lens? **Draw a ray diagram** to justify your answer. A 4 cm tall object is placed perpendicular to its principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 15 cm. Find the** nature, position, and size **of the image.

One-half of a convex lens of focal length 10 cm is covered with black paper. Yes, such a lens can produce an image of a complete object placed at a distance of 30 cm from the lens. Because the light coming from the object can be refracted from the other half of the lens. The image formed will be **real, inverted, **and **diminished.**

The ray diagram for the justification of the answer is as follows:

**Given:**

Object size, $h=+4cm$

Object distance, $u=–15cm$

Focal length, $f=+20cm$

**
**

**To find:** Image distance, or position $v$, nature and size of the image, $h'$.

**
**

**Solution:**

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

Substituting the required values, we get-

$\frac {1}{20}=\frac {1}{v}-\frac {1}{(-15)}$

$\frac {1}{20}=\frac {1}{v}+\frac {1}{15}$

$\frac {1}{v}=\frac {1}{20}-\frac {1}{15}$

$\frac {1}{v}=\frac {3-4}{60}$

$\frac {1}{v}=-\frac {1}{60}$

$v=-60cm$

Thus the screen image distance or position, $v$ is **60 cm** from the lens, and the minus sign implies that the image forms in front of the lens (on the left side). Also, the image will be **virtual **as it forms on the left side of the lens.

Now,

From the magnification formula, we know that-

$m=\frac {v}{u}=\frac {h'}{h}$

Substituting the required values, we get-

$\frac {-60}{-15}=\frac {h'}{4}$

$4=\frac {h'}{4}$

$h'=+16cm$

Thus, the height or size of the image, $h'$ is 16 cm, and the plus sign implies that the image forms **erect (**upright **)**.

Therefore, the image formed will be** virtual, erect, **and** magnified.**