Four transparent planes A, B, C and D are shown in front view on the left. But when viewed in perspective, they appear like the image shown on the right. Which plane is the farthest from the viewer?

Question

Coloured strips are printed on a transparent sheet as shown below. The bottom half of this sheet is folded backwards along the PQ axis to form a semi-circle. Which of the option(s) will be a part of the resultant semi-circle?

Answer 1

Try each plane as the "farthest" one and check if the nested picture still makes sense.

D: If D were farthest, it would have to be the outer ring in the nested view, since a plane placed farther back has more of the scene between it and the viewer and needs to be shown as the outer boundary. But D is drawn as the innermost ring, so this does not work.
C: The same problem shows up for C. If C were farthest, it should appear as the outer ring, but it is drawn one ring in from B, not on the outside. This option fails too.
B: B sits one ring in from the outer boundary, so treating it as the farthest plane would again clash with its position in the nested drawing.
A: A is the only plane drawn as the outer ring. Placing A farthest back matches its position perfectly, since the plane at the back should be the one framing all the others in the nested view.

Only option A is free of contradictions once we line up "farthest back" with "outer ring" in the nested picture.

So the correct answer is A.

Answer 2

Try each plane as the "farthest" one and check if the nested picture still makes sense.

D: If D were farthest, it would have to be the outer ring in the nested view, since a plane placed farther back has more of the scene between it and the viewer and needs to be shown as the outer boundary. But D is drawn as the innermost ring, so this does not work.
C: The same problem shows up for C. If C were farthest, it should appear as the outer ring, but it is drawn one ring in from B, not on the outside. This option fails too.
B: B sits one ring in from the outer boundary, so treating it as the farthest plane would again clash with its position in the nested drawing.
A: A is the only plane drawn as the outer ring. Placing A farthest back matches its position perfectly, since the plane at the back should be the one framing all the others in the nested view.

Only option A is free of contradictions once we line up "farthest back" with "outer ring" in the nested picture.

So the correct answer is A.

Answer 3

A basic rule of vision explains this picture: an object looks smaller the farther it sits from the viewer, and for two objects to appear one inside the other from a single spot, the one drawn on the outside has to occupy the larger real span even though it sits deeper in the scene. Let's use this rule on each plane.

A: A takes up the outer, widest span in the nested picture. For a plane to still look this wide from the viewer's spot despite being pushed back in depth, it must be the plane placed farthest away.
B: B's span is smaller than A's, so by the same rule it does not need to sit as deep as A to produce that smaller look, placing it nearer than A.
C: C's span shrinks again, so it sits nearer than B by the same logic.
D: D has the smallest span of all four, so it needs the least depth to look that small, putting it closest to the viewer.

Since apparent size falls as distance grows, and A is the one that keeps the largest apparent size while still being pushed to the back of the arrangement, A must be the plane placed farthest from the viewer.

So the correct answer is A.

Answer 4

A basic rule of vision explains this picture: an object looks smaller the farther it sits from the viewer, and for two objects to appear one inside the other from a single spot, the one drawn on the outside has to occupy the larger real span even though it sits deeper in the scene. Let's use this rule on each plane.

A: A takes up the outer, widest span in the nested picture. For a plane to still look this wide from the viewer's spot despite being pushed back in depth, it must be the plane placed farthest away.
B: B's span is smaller than A's, so by the same rule it does not need to sit as deep as A to produce that smaller look, placing it nearer than A.
C: C's span shrinks again, so it sits nearer than B by the same logic.
D: D has the smallest span of all four, so it needs the least depth to look that small, putting it closest to the viewer.

Since apparent size falls as distance grows, and A is the one that keeps the largest apparent size while still being pushed to the back of the arrangement, A must be the plane placed farthest from the viewer.

So the correct answer is A.