A biconvex lens of focal length f1=10 cm is placed 40 cm in front of a concave mirror of focal length f2=7.50 cm, as shown in the figure. An object, 2 cm high, is placed 20 cm to the left of the lens. The image formed by the lens, as rays travel to the right, is

Question

A biconvex lens of focal length f1=10 cm is placed 40 cm in front of a concave mirror of focal length f2=7.50 cm, as shown in the figure. An object, 2 cm high, is placed 20 cm to the left of the lens. <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-or92elz5hd7ujlys.jpg" /> The image formed by the lens, as rays travel to the right, is (A) Real and erect (B) Virtual and erect (C) Real and inverted (D) Virtual and inverted

Tardigrade Admin · Accepted Answer

From lens equation, (1/v)-(1/(- 20))=(1/10) ⇒ v=+20cm Magnification, m1=(v/u)= ((+ 20/- 20) )=-1 Image is real, inverted, and same size as object.

Q. A biconvex lens of focal length $f_{1} = 10 c m$ is placed $40 c m$ in front of a concave mirror of focal length $f_{2} = 7.50 c m,$ as shown in the figure. An object, $2 c m$ high, is placed $20 c m$ to the left of the lens.

The image formed by the lens, as rays travel to the right, is

Real and erect

Virtual and erect

Real and inverted

Virtual and inverted

From lens equation,
$\frac{1}{v} - \frac{1}{( - 20 )} = \frac{1}{10} \Rightarrow v = + 20 c m$
Magnification, $m_{1} = \frac{v}{u} = (\frac{+ 20}{- 20}) = - 1$
Image is real, inverted, and same size as object.

Q. A biconvex lens of focal length f1​=10cm is placed 40cm in front of a concave mirror of focal length f2​=7.50cm, as shown in the figure. An object, 2cm high, is placed 20cm to the left of the lens. The image formed by the lens, as rays travel to the right, is