Ray Optics And Optical Instruments - Class 12 Medical Physics - Extra Questions

• Two times a ray of light undergo refraction when it passes through a water droplet.
  
• One time at air-water interface and one time at water-air interface.
  

Since the critical angle of glass is $${42°}$$(given). So,we will be observing total internal reflection. Now we will be observing inverted image in the prism at same disrance in parallel to the lens.

1.  The focal length of the eyepiece of a compound microscope is larger so that the final image formed is a highly magnified image.
2. A telescope of a binocular has two total internal reflecting prisms which help in inverting the image formed by the objective as a  result the final image is erect and magnified.

Focal length of objective $$=  f_0$$
Focal length of eyepiece$$ = f_e $$
Magnification $$=\dfrac{ f_0}{f_ e}$$.
 When the focal length of eyepiece decreases the magnification increases. We get a bigger and clearer image of the object viewed.

Telescopes are  used to obtain an enlarged image of a distant object, it uses a curved lens or mirror to collect light from distant objects and to focus that light on an image and creates a brighter and sharper image. The focused image formed by the objective lens of a telescope is magnified by a smaller second lens called an eyepiece.  

A magnifying glass is a concave mirror which is used to produce a magnified image of an object. The lens is usually mounted in a frame with a handle.

It produces three times magnified real image of an object. The image formed by a concave mirror is observed to be virtual, erect and larger than the object.

Telescope is a  optical devices which uses two lenses, primary lens forms a real image and a secondary lens is used as a magnifier to make an enlarged virtual image.When an object is placed farther from a convergent lens than the focal length, the lens produces an inverted, real image and enlarged image, depending how close the object is to the focal point.

Given that,

Focal length $$f=\dfrac{R}{2}$$

Radius of curvature $$R= 30$$

Object distance $$u=10\,cm$$

Height of object $${{h}_{o}}=2\,cm$$

From the mirror formula,

  $$ \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u} $$

 $$ -\dfrac{1}{v}=\dfrac{1}{u}-\dfrac{1}{f} $$

 $$ -\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{2}{30} $$

 $$ -\dfrac{1}{v}=\dfrac{3-2}{30} $$

 $$ v=-30\,cm $$

Now, the magnification is

  $$ m=\dfrac{-v}{u} $$

 $$ m=\dfrac{30}{10} $$

 $$ m=3 $$

 Now, the height of image

  $$ m=\dfrac{{{h}_{i}}}{{{h}_{o}}} $$

 $$ 3=\dfrac{{{h}_{i}}}{2} $$

 $$ {{h}_{i}}=6\,cm $$

 Hence, the image distance is $$-30\ cm$$, height of image is $$6\ cm$$ and magnification is $$3$$

By applying the lens maker formula

$$\frac{1}{f} = \left( {n - 1} \right)\left( {\frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}} \right)$$

The focal length of the lens will be

$$\left( {n - 1} \right) = \left( {\frac{2}{3}\frac{{{R_1}}}{f}} \right)$$

Hence the refractive index will be$$n = 1 + \frac{2}{3}\frac{{{R_1}}}{f}$$

For first lens

$$ {{f}_{1}}=+10\,cm $$

$$ {{u}_{1}}=\,-30\,cm $$

Using lens formula

$$ \dfrac{1}{{{v}_{1}}}-\dfrac{1}{{{u}_{1}}}=\dfrac{1}{{{f}_{1}}} $$

$$ \dfrac{1}{{{v}_{1}}}=\dfrac{1}{{{f}_{1}}}+\dfrac{1}{{{u}_{1}}} $$

$$ {{v}_{1}}=+15\,cm $$

For the first lens image is formed at a distance of $$15\,cm$$ from the right of it. This image behaves as object for second lens.

 For second lens

$$ {{f}_{2}}=-10\,cm $$

$$ {{u}_{2}}=15-5=+10\,cm $$

$$ \dfrac{1}{{{v}_{2}}}=\dfrac{1}{{{f}_{2}}}+\dfrac{1}{{{u}_{2}}} $$

$$ \dfrac{1}{{{v}_{2}}}=\dfrac{-1}{10}+\dfrac{1}{10} $$

$$ {{v}_{2}}=\infty  $$

Real image is formed by the second lens at infinite distance. This image will acts as an object for third lens.

 For third lens

$$ {{f}_{3}}=+30\,cm $$

$$ {{u}_{3}}=\infty  $$

$$ \dfrac{1}{{{v}_{3}}}=\dfrac{1}{{{f}_{3}}}+\dfrac{1}{{{u}_{3}}} $$

$$ \dfrac{1}{{{v}_{3}}}=\dfrac{1}{30}+\dfrac{1}{\infty } $$

$$ {{v}_{3}}=+30\,cm $$

So, the final image is formed at distance of 30 cm to the right of third lens.

Refraction at Spherical Surface

Distance of object and image from spherical pole is $$u\,\,and\,\,v$$

Redfractive index of rarer medium and denser medium is $${{\mu }_{1}}\,\,and\,\,{{\mu }_{2}}$$

We get,

$$\tan \,\alpha =\dfrac{MN}{OM},\,\,\tan \beta =\dfrac{MN}{MI}\,\,\,and\,\,\tan \gamma =\dfrac{MN}{MC}$$

Now, for Δ NOC

$$i\text{ }=~\angle NOM\text{ }+\angle NCM$$

$$i=\dfrac{MN}{OM}+\dfrac{MN}{MC}$$ …….1

Similarly,

$$r=\dfrac{MN}{MC}\text{ +}\dfrac{MN}{MI}$$ …….2

Now by using Snell’s law we get

$${{\mu }_{1}}sin\text{ }i\text{ }=\text{ }{{\mu }_{2}}si{{n}_{~}}r$$

Substituting i and r from Eq. (1) and (2), we get

$$\dfrac{{{\mu }_{1}}}{OM}+\dfrac{{{\mu }_{2}}}{MI}=\dfrac{{{\mu }_{2}}-{{\mu }_{1}}}{MC}$$

As $$OM=-u,\text{ }\,MI=+v,\text{ }\,MC=+R$$

Hence, the equation becomes, $$\dfrac{{{\mu }_{2}}}{v}\text{-}\dfrac{{{\mu }_{1}}}{u}=\dfrac{{{\mu }_{2}}-{{\mu }_{1}}}{R}$$

We know that,

Refractive Index of the glass$${{\mu }_{1}}=1.5$$

Refractive Index of the air $${{\mu }_{2}}=1$$

Radius of the Convex surface $$r=-15\,cm$$

Real Depth of the Air Bubble $$u=-6\,cm$$

Let the Apparent Depth of the Air bubble $$=v\,cm$$

By using formula  for refraction on the curved surface

  $$ \dfrac{{{\mu }_{2}}}{v}-\dfrac{{{\mu }_{1}}}{u}=\dfrac{{{\mu }_{2}}-{{\mu }_{1}}}{R} $$

 $$ \dfrac{1}{v}-\dfrac{1.5}{-6}=\dfrac{1-1.5}{-15} $$

 $$ \dfrac{1}{v}=\dfrac{0.5}{-15}+\dfrac{1.5}{-6} $$

 $$ \dfrac{1}{v}=\dfrac{-1}{30}-\dfrac{1}{4} $$

 $$ v=\dfrac{-120}{34} $$

 $$ v=\dfrac{-60}{17} $$

 $$ v=-3.5\,cm $$

• The parallel rays incident on convex lens coveergae at focus F of lens .
  
• At mirror,after reflection forms an image at optical center of lens C.
  
• Thus image of virtual point object F is formes at c.
  

1.From the observation 3, we can say that Radius of curvature is 40cm so foal length of the convex lens is 20cm.[f=R/2]

2. Observation 6 is incorrect , as the object is in between P and F, the image formed will be virtual image on the same side of object, image distance should be negative.

3.Ray diagram for observation 3: Please refer attachment : object beyond C

4. magnification=m=v/um=30/-60=-1/2= -0.5so the image is real, inverted and diminished.