Apparent depth of mark as seen through a glass slab of thickness $$x$$ and refractive index $$\mu $$ is
Apparent depth $$ = \frac{{{\text{real depth}}}}{{{\text{refractive index}}}}$$
$${\text{or}}\,\,x' = \frac{x}{\mu } = \frac{3}{{1.5}} = 2\,cm$$
As image appears to be raised by $$1\,cm,$$  therefore, microscope must be moved upward by $$1\,cm.$$

Let the object distance be $$x.$$ Then, the image distance is $$D - x.$$
From lens equation, $$\frac{1}{x} + \frac{1}{{D - x}} = \frac{1}{f}$$
On algebraic rearrangement, we get
$$x2 - Dx + Df = 0$$
On solving for $$x,$$ we get
$$\eqalign{ & {x_1} = \frac{{D - \sqrt {D\left( {D - 4f} \right)} }}{2} \cr & {x_2} = \frac{{D + \sqrt {D\left( {D - 4f} \right)} }}{2} \cr} $$
The distance between the two object positions is
$$d = {x_2} - {x_1} = \sqrt {D\left( {D - 4f} \right)} $$

For the image of point $$P$$ to be seen by the observer, it should be formed at point $$Q.$$
In $$\Delta QNS,$$
$$\eqalign{ & NS = QS = 2\,h \cr & \therefore \,\,\angle NQS = {45^ \circ } \cr & \therefore \,\,r = {45^ \circ } \cr & {\text{Now in }}\Delta QMA, \cr & \angle MQA = {45^ \circ } \cr & \therefore \,MA = QA = h \cr & _2^1\mu = \frac{{\sin r}}{{\sin i}} = \frac{{\sin {{45}^ \circ }}}{{\sin i}} \cr & {\text{In }}\Delta PMB, \cr & P{M^2} = 4\,{h^2} + {h^2} \cr & = 5\,{h^2} \cr} $$
$$\therefore \,\,\sin i = \frac{h}{{\sqrt 5 h}} = \frac{1}{{\sqrt 5 }}\,\,\,\,.....\left( {{\text{ii}}} \right)$$
From (i) and (ii)
$$_2^1\mu = \sqrt {\frac{5}{2}} $$

Frequency remains constant during refraction
$$\eqalign{ & {v_{med}} = \frac{1}{{\sqrt {{\mu _0}{ \in _0} \times \,4} }} = \frac{c}{2} \cr & \frac{{{\lambda _{med}}}}{{{\lambda _{air}}}} = \frac{{{v_{med}}}}{{{v_{air}}}} = \frac{{\frac{c}{2}}}{c} = \frac{1}{2} \cr} $$
∴ wavelength is halved and frequency remains unchanged

Cutting a lens in transverse direction doubles their focal length i.e. $$2f.$$
Using the formula of equivalent focal length,
$$\frac{1}{f} = \frac{1}{{{f_1}}} + \frac{1}{{{f_2}}} + \frac{1}{{{f_3}}} + \frac{1}{{{f_4}}}$$
We get equivalent focal length as $$\frac{f}{2}.$$

In any medium other than air or vacuum, the velocities of different colours are different. Therefore, both red and green colours are refracted at different angles of refraction. Hence, after emerging from glass slab through opposite parallel face, they appear at two different points and move in the two different parallel directions.

According to Cauchy relation,
$$\eqalign{ & \mu = A + \frac{B}{{{\lambda ^2}}} + \frac{C}{{{\lambda ^4}}} + \ldots \cr & \Rightarrow \mu \propto \frac{1}{\lambda } \cr} $$
Hence, focal length $$f \propto \lambda $$
For a convex lens, $${f_R} > {f_V}\,\,{\text{or}}\,\,{f_V} < {f_R}.$$
For a concave lens, focal length is negative. So, focal length of concave lens for violet light $${F_V} > $$  Focal length of concave lens for red light $${F_R}.$$

For dispersion without deviation, net deviation produced by the combination of prisms must be zero.
Let, prism angle of the first and second prisms are $${A_1}$$ and $${A_2}$$ respectively. Similarly, their refractive indices are $${\mu _1}$$ and $${\mu _2}.$$
Condition for dispersion without deviation is
$$\eqalign{ & {\delta _1} - {\delta _2} = 0 \cr & \Rightarrow \left( {{\mu _1} - 1} \right){A_1} - \left( {{\mu _2} - 1} \right){A_2} = 0 \cr & \Rightarrow {A_2} = \left( {\frac{{{\mu _1} - 1}}{{{\mu _2} - 1}}} \right) \cr & {A_1} = \left( {\frac{{1.42 - 1}}{{1.7 - 1}}} \right)\left( {{{10}^ \circ }} \right) \cr & \Rightarrow {A_2} = {6^ \circ } \cr} $$

See geometry of the figure.

Huygen gave a hypothesis for geometrical construction of the position of a common wavefront at any instant, during the propagation of waves in a medium. Huygen’s principle can be used to explain the phenomenon of reflection, refraction and diffraction of light. Origin of spectra is explained by quantum theory.