Physics Ray Optics and Optical Instruments

dist l1 l2 $=2\text{exception 25:}a$

= 2 * 2.3 a

= 4.6 a

$\frac{1}{f_1}=\left({\mu }_1-1\right)\left(\frac{1}{\infty }-\frac{1}{\left(-R\right)}\right)=\frac{{\mu }_1-1}{R}$

$\frac{1}{f_2}=\left({\mu }_2-1\right)\left[\frac{1}{\left(-R\right)}-\frac{1}{\infty }\right]=-\frac{{\mu }_2-1}{R}$

$\frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{{\mu }_1-{\mu }_2}{R}$

$\Rightarrow \frac{R}{f_{eq}}={\mu }_1-{\mu }_2$

$\frac{1}{V_1}+\frac{1}{30}=\frac{1}{10}$

V₁ = 15 cm

$\frac{1}{V_2}-\frac{1}{10}=-\frac{1}{10}$

$\Rightarrow V_2=\infty$

$\Rightarrow V_3=30cm$

$\Rightarrow O{V}_3=75cm$

As image coincides with the object, image formed by convex lens should be at the centre of curvature of the convex mirror. If the convex mirror is removed, then, also, image formed the convex lens is at the same location.

Distance of image from object = 12 + 8 + 2 * 15 = 50 cm

For mirror

${\overrightarrow{V}}_{l_{1}M}=-m^2{\overrightarrow{V}}_{O_{1}M}$

$$ ${\overrightarrow{V}}_{l_{1}M}$  Velocity of image w.r.t. mirror

$$ ${\overrightarrow{V}}_{0M}$  Velocity of object w.r.t. mirror

m magnification

$\frac{1}{v}+\frac{1}{-1.9}=\frac{1}{-0.1}$

$\frac{1}{v}=\frac{1}{1.9}-\frac{1}{0.1}$

$=\frac{0.1-1.9}{1.9*0.1}$

$v=\frac{1.9*0.1}{-1.8}=\frac{0.19}{-1.8}=-0.105$

$m=-\frac{v}{u}=-\left(\frac{-0.1}{-1.9}\right)$

$\left|{\overrightarrow{V}}_{l_{1}M}\right|={\left(\frac{1}{19}\right)}^2\left|{\overrightarrow{V}}_{O_{1}M}\right|$

$=\frac{1}{{\left(19\right)}^2}*40=0.1m/s$

${\mu }_B=\frac{C}{v_B}$

$v_B=\frac{3*10^8}{1.47}=20.408*10^7$

$v_A=2.6*10^7+20.408*10^7=23*10^7$

${\mu }_B=\frac{C}{v_B}\&{\mu }_A=\frac{C}{v_A}$

= 1.127

$\approx 1.13$

Velocity of light in medium 2

$=\frac{1}{\sqrt[]{{\mu }_m{\in }_m}}$

$v_2=\frac{1}{\sqrt[]{{\mu }_0{\mu }_r{\in }_0{\in }_r}}$

$=\frac{1}{\sqrt[]{1*{\mu }_0{\in }_0*4}}$

$v_2=\frac{1}{2\sqrt[]{{\mu }_0{\in }_0}}$

By snell's law for total internal Reflection

${\mu }_{2}sin{\theta }_C>{\mu }_{1}sin90°$

${\theta }_c>30°$

Using snell's law –

sin I = $\mu sinr$

sin45° = $\mu sin30°$

$\mu$$=$$\frac{sin45}{sin30}$$=$$\frac{1}{\text{exception 25:}}$$*$$2$$=$√2

The Class 12 Physics Ch 10 Wave Optics includes a significant weightage in CBSE Class 12 Physics Exams which carries around 10 to 12 marks. Topics like Diffraction, Polarization, and Young's Double-Slit Experiment are oftenly asked in theory-based and numerical questions in CBSE Board exams and other competitive exams and therefore it is an important chapter for exams preparation.