Question

Consider a resonator with one plane mirror and other with radius =1 m, the distance between two resonators is 75 cm. Given λ0=1µm. Obtain the transverse intensity distributions of the fundamental mode at the position of the two mirrors.​

Answer 1

Explanation:

Given that:

QP=8cm,PR=6cm and SR=3cm

(I) In △PQR and △SPR

∠PRQ=∠SRP (Common angle)

∠QPR=∠PSR (Given that)

∠PQR=∠PSR (Properties of triangle )

∴△PQR∼△SPR (By AAA)

(II)

SP

PQ

=

PR

QR

=

SR

PR

(Properties of similar triangles)

⇒

SP

8cm

=

3cm

6cm

⇒SP=4cm and

⇒

6cm

QR

=

3cm

6cm

⇒QR=12cm

(III)

ar(△SPR)

ar(△PQR)

=

SP

2

PQ

2

=

4

2

8

2

=