Newton\'s rings. The figure shows how fringes of equal thickness may be produced

Question

Newton's rings. The figure shows how fringes of equal thickness may be produced (Newton's rings). A spherical surface of index n1 rests on a flat surface of the same index. A gap of variable thickness and index n2 produces the interference fringes. For constructive interference, give the dependence of d on m and Lambda. Calculate rm , the radius of the mth fringe as a function of m, Lambda, R4 and n2 and use the approximation R>>d. For Lambda = 0.5 mu m, n2 = 1, and R = 5m, calculate the radius of the first two rings.

Explanation / Answer

a) 2d=(2m+1)*lamda/2

b)Radii of the mth dark rings: rm=(m*lamda*R)0.5

Radii of the mth bright ring: rm=((2m+1)*R*lamda/2)0.5

c)center will be dark.so radius of ist fringe will be isr bright ie r1=(m*lamda*R)0.5=(0.5*5)0.5=1.58m

second will be 1sst dark r2=((2m+1)*R*lamda/2)0.5=((3)*5*.5/2)0.5=1.936m

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.