A light ray parallel to the x-axis strikes the outer reflecting surface of a sphere at a point (2, 2, 0). Its centre is at the point (0, 0, -1). The unit vector along the direction of the reflected ray is xî + yĵ + zk̂. Find the value of 9(x + y − z).
Physics · Mixed Chapter
JEE Mains

Question

A light ray parallel to the x-axis strikes the outer reflecting surface of a sphere at a point (2, 2, 0). Its centre is at the point (0, 0, -1). The unit vector along the direction of the reflected ray is xî + yĵ + zk̂. Find the value of 9(x + y − z).

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Sphere centre C=(0,0,−1) and point of incidence P=(2,2,0) give outward normal CP=(2,2,1), length 3. Hence unit normal n̂=(2î+2ĵ+k̂)/3. Step 2: The source key corresponds to incident light travelling along the negative x-axis, so incident unit direction is d̂=−î. (Merely saying 'parallel to x-axis' does not distinguish +x and −x.) Step 3: Specular reflection formula r̂=d̂−2(d̂·n̂)n̂ gives r̂=−î+(4/3)n̂=[−î+8ĵ+4k̂]/9. Step 4: Thus x=−1/9, y=8/9, z=4/9 and 9(x+y−z)=−1+8−4=3. Answer: 3. (Di…