A convex mirror has radius of curvature of 20 cm. An object is placed at such a distance from the mirror that the size of image is exactly half that o

Correct option is (d) Between 0 cm and 15 cm

(a) Using the relation, `(1)/(u)+(1)/(v)=(1)/(f)` `(1)/(v)=(1)/(f)-(1)/(u)=(1)/(-18)=(1)/((-27))` v=-54cm The negative sign shows that the image is formed in front of the mirror, i.e., on the side of the object itself. Thus,...

Given,R=-24 cm Hence, `f=R/2=-12 cm` (i) image is virtual and three times larger. Hence, u is negative and v is positive. Simultaneousely, `abs(v)=3abs(u)`. So let, u=-x then, v=+3x Substituting in...

Correct Answer - D When object lies between focs and pole of convex lens, image is virtual, erect and enlarged.

Correct Answer - C (c ) We have, `I/O=f/((f-u))` `rArr I/5=(-10)/(-10-(-100))` Size of the image , I=0.55 cm

Correct Answer - D For virtual object (u = positive), `1/v+1/u=1/f` `therefore 1/v=1/f-1/u` If `vltf`, then v=-ve i.e., image may be real.

Correct Answer - C (c) Glass slab will shift the object towards mirror. If it comes between pole and focus the image will become virtual.

Correct Answer - D Focal length of a convex lens by displacement method is given as, `f=(a^(2)-b^(2))/(4a)` Where, a= distance between the image and object and b=distance between two position of...

Correct Answer - D For the best first condition, `f_(1)=10cm u=-30cm` then, `(1)/(f_(1))=(1)/(v)-(1)(u) Rightarrow=(1)/(v)+(1)/(30)` `(1)/(v)=(1)/(10)-(1)/(30) Rightarrow v=(30)/(2)=15` For the second condition, when concave lens is placed, (where F=focal length of combination)...

Correct Answer - A Given `f=-f Rightarrow v=(1)/(3)u` According to the lens formula, `(1)/(f)=(1)/(v)-(1)/(u)-(1)/(f)=(1)/((-1//2)u)+(1)/(u)-(1)/(f)` `(-3+1)/(u)-(1)/(f)=-(2)/(u) Rightarrow u=2f`