One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

 From the lens formula, we know that-

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

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.