Physics - Geometrical Optics Question with Solution | TestHub

$(v_{i/m}{)}_x=-(v/u{)}^2(v_{0/m}{)}_x$

$(v_i{)}_x=-(v/u{)}^2((v_0{)}_x-v_m)+v_m$

In the frame of the mirror vs the frame of the ground: if $(v_o{)}_x>>v_m$ then $(v_i{)}_x$ is in the backward direction, and if $(v_o{)}_x<<v_m$ then $(v_i{)}_x$ is in the forward direction.

$(v_i{)}_y=(dm/dt)y_0=[d/dt(-v/u)]y_0$

If the object is approaching the focal point (F), then m increases and |y_i| also increases; if the object is approaching the pole, then m decreases and |y_i| decreases.

(P) $v_1<v_2$ : In the frame of mirror:

In the frame of ground: if $(v_i{)}_x>>v_1$ :

option (C)

If $(v_i{)}_x<<v_1$ :

option (D)

$v_1>v_2$ : In the frame of mirror:

In the ground frame:

option (A)

(Q) In the frame of mirror:

In the ground frame:

option (A)

(R):

option (B)

(S):

option (D)

(T) $V_1>V_2$ : In the frame of mirror:

In the frame of ground: if $(v_i{)}_x<v_2$ :

option (C)

If $(v_i{)}_x>v_2$ :

option (D)

$V_1<V_2$ : In the frame of mirror:

In the ground frame:

option (B)

Hence: (A) P,Q (B) R,T (C) P,T (D) P,S,T.