An object ' o ' is placed at a distance of 100 cm in front of a concave mirror of radius of curvature 200 cm as shown in the figure. The object starts moving towards the mirror at a speed 2 cm / s. The position of the image from the mirror after 10 s will be at ....... cm.

Question

An object ' o ' is placed at a distance of 100 cm in front of a concave mirror of radius of curvature 200 cm as shown in the figure. The object starts moving towards the mirror at a speed 2 cm / s. The position of the image from the mirror after 10 s will be at ....... cm. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

After 10 sec. u =-80 cm f =-100 cm (1/ v )+(1/ u )=(1/ f ) v =400 cm

Q. An object ' $o$ ' is placed at a distance of $100 c m$ in front of a concave mirror of radius of curvature $200 c m$ as shown in the figure. The object starts moving towards the mirror at a speed $2 c m / s$ . The position of the image from the mirror after $10 s$ will be at ....... $c m$ .

After $10 sec$ .
$u = - 80 c m$
$f = - 100 c m$
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$v = 400 c m$

Q. An object ' o ' is placed at a distance of 100cm in front of a concave mirror of radius of curvature 200cm as shown in the figure. The object starts moving towards the mirror at a speed 2cm/s. The position of the image from the mirror after 10s will be at ....... cm.

After 10sec. u=−80cm f=−100cm v1​+u1​=f1​ v=400cm

Q. An object ' $o$ ' is placed at a distance of $100 c m$ in front of a concave mirror of radius of curvature $200 c m$ as shown in the figure. The object starts moving towards the mirror at a speed $2 c m / s$ . The position of the image from the mirror after $10 s$ will be at ....... $c m$ .

After $10 sec$ .
$u = - 80 c m$
$f = - 100 c m$
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$v = 400 c m$