Answer:
θ₂ = 3 θ₁
Explanation:
given,
telescope of lens diameter = 12 inch
another telescope of lens diameter = 4 inch
comparison of resolution power.
Using the formula of resolution
[tex]\theta = \dfrac{1.22 \lambda}{D}[/tex]
for diameter = 12 inch
[tex]\theta_1 = \dfrac{1.22 \lambda}{D_1}[/tex].....(1)
for diameter = 4 inch
[tex]\theta_2 = \dfrac{1.22 \lambda}{D_2}[/tex].......(2)
dividing equation (2) from (1)
[tex]\dfrac{\theta_2}{\theta_1} = \dfrac{D_1}{D_2}[/tex]
now,
[tex]\dfrac{\theta_2}{\theta_1} = \dfrac{12}{4}[/tex]
[tex]\dfrac{\theta_2}{\theta_1} =3[/tex]
θ₂ = 3 θ₁
hence, we can say that resolution of telescope of 12 inch is 3 time smaller than the resolution of 4 inch telescope.
