A concave mirror of focal length 1.5m forms an image of an object plac

1.	A concave mirror of focal length 1.5m forms an image of an object placed at a distance of 40cm. Find the position and nature of the image.
Answer» According to the question; Focal length (f) = -1.5m; Object distance (u) = -40cm = 0.40m; (1m = 100 cm) By mirror formula; \(\frac1v+\frac1u=\frac1f\) ⇒ \(\frac1v+\frac1{-0.40}=\frac1{-1.5}\) ⇒ \(\frac1v=\frac1{0.40}-\frac1{1.5}\) ⇒\(\frac1v=\frac{10}{4}-\frac{10}{15}\) ⇒ \(\frac1v=\frac{5}{2}-\frac{2}{3}\) ⇒ \(\frac1v=\frac{15-4}{6}\) ⇒ \(\frac{1}{v}=\frac{11}{6}\) ⇒ v = 0.5454m = 54.54cm (1m = 100 cm). Since v = 54.54cm which is positive hence image is behind the mirror. The image is formed at the distance of 54.54 cm behind the mirror and is virtual and erect.

A concave mirror of focal length 1.5m forms an image of an object placed at a distance of 40cm. Find the position and nature of the image.

Answer»

According to the question;

Focal length (f) = -1.5m;

Object distance (u) = -40cm = 0.40m; (1m = 100 cm)

By mirror formula;

\(\frac1v+\frac1u=\frac1f\)

⇒ \(\frac1v+\frac1{-0.40}=\frac1{-1.5}\)

⇒ \(\frac1v=\frac1{0.40}-\frac1{1.5}\)

⇒\(\frac1v=\frac{10}{4}-\frac{10}{15}\)

⇒ \(\frac1v=\frac{5}{2}-\frac{2}{3}\)

⇒ \(\frac1v=\frac{15-4}{6}\)

⇒ \(\frac{1}{v}=\frac{11}{6}\)

⇒ v = 0.5454m = 54.54cm (1m = 100 cm).

Since v = 54.54cm which is positive hence image is behind the mirror.

The image is formed at the distance of 54.54 cm behind the mirror and is virtual and erect.