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NCERT Exemplar · Q31

Q.An infinitely long solid cylinder of radius RR is made of an unusual exotic material whose refractive index is exactly −1-1. The cylinder is placed between two parallel horizontal planes (a lower plane and an upper plane) whose normals point in the yy direction; the axis and centre O of the cylinder lie on the yy-axis, midway, with the cylinder's circular cross-section in the xyxy-plane. A narrow laser beam is fired straight upward (in the +y+y direction) from the lower plane, at a horizontal distance xx from the vertical diameter of the cylinder (so xx is the perpendicular distance of the beam from the axis, i.e. its impact parameter). Find the range of xx for which the light emitted from the lower plane fails to reach the upper plane.

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For a medium of index −1-1 the refracted ray sits on the same side of the normal as the incident ray, so each refraction turns the ray by 2i2i; entry and exit together turn it by 4i4i. The upward beam therefore emerges rotated by 4i4i from the vertical, and it can no longer climb to the upper plane once 4i4i exceeds 90∘90^\circ (until it swings past 270∘270^\circ). This gives 22.5∘<i<67.5∘22.5^\circ<i<67.5^\circ, i.e. Rsin⁡22.5∘<x<Rsin⁡67.5∘R\sin22.5^\circ<x<R\sin67.5^\circ.

Incidence angle

The beam travels along +y+y and strikes the lower surface of the cylinder. With impact parameter xx, the radius to the strike point makes the normal, and the angle of incidence ii satisfies

sin⁡i=xR.\sin i=\frac{x}{R}.

(For x>Rx>R the beam misses the cylinder entirely and goes straight to the upper plane.)

Negative refraction: deviation at each surface

Snell's law with n2=−1n_2=-1 gives sin⁡i=∣n2∣sin⁡r=sin⁡r\sin i=|n_2|\sin r=\sin r, so ∣r∣=i|r|=i, but the refracted ray lies on the same side of the normal as the incident ray (the negative-index signature). Resolving the ray into components along and across the normal, the tangential component reverses while the normal component keeps going forward. Starting from the upward direction (0,1)(0,1), the ray inside the cylinder becomes

(−sin⁡2i, cos⁡2i),(-\sin 2i,\ \cos 2i),

i.e. it is rotated by 2i2i. Tracing the chord to the far surface and refracting again (by symmetry the second incidence angle is also ii), the emergent ray direction works out to

(−sin⁡4i, cos⁡4i).(-\sin 4i,\ \cos 4i).

So the total deviation is 4i4i.

Condition for not reaching the upper plane

The emergent ray has vertical (upward) component cos⁡4i\cos 4i. It can climb to the upper plane only if this is positive. Hence the light fails to reach the top when

cos⁡4i<0  ⟹  90∘<4i<270∘  ⟹  22.5∘<i<67.5∘.\cos 4i<0 \;\Longrightarrow\; 90^\circ<4i<270^\circ \;\Longrightarrow\; 22.5^\circ<i<67.5^\circ.

Convert to xx

Using x=Rsin⁡ix=R\sin i: …

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