Refraction Of Light Ray Optics & Optical Instruments

Question 1:

A monochromatic light ray is incident at an angle of $60^\circ$ on the surface of a glass slab of refractive index $\sqrt{3}$. The angle of refraction inside the glass slab is $r$. What is the value of $r$?

(1) $30^\circ$ (2) $45^\circ$ (3) $60^\circ$ (4) $\sin^{-1}(\frac{1}{\sqrt{3}})$

Solution:

According to Snell’s Law, the relationship between the angle of incidence ($i$), the angle of refraction ($r$), and the refractive indices of the two media ($n_1$ and $n_2$) is given by:

$$n_1 \sin i = n_2 \sin r$$

Here, the light ray is incident from air, so $n_1 = 1$. The refractive index of the glass slab is $n_2 = \sqrt{3}$, and the angle of incidence is $i = 60^\circ$. Plugging these values into Snell’s Law:

$$1 \cdot \sin 60^\circ = \sqrt{3} \sin r$$$$\frac{\sqrt{3}}{2} = \sqrt{3} \sin r$$$$\sin r = \frac{\sqrt{3}}{2 \sqrt{3}} = \frac{1}{2}$$ $$r = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ$$

Therefore, the correct answer is (1) $30^\circ$.

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Question 2:

A compound microscope has an objective lens with a focal length of 2.0 cm and an eyepiece with a focal length of 5.0 cm. An object is placed 2.5 cm from the objective lens. If the final image is formed at the least distance of distinct vision (D = 25 cm), what is the magnifying power of the microscope?

(1) 12.5 (2) 25 (3) 100 (4) 250

Solution:

First, let’s find the image distance ($v_o$) formed by the objective lens using the lens formula:

$$\frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o}$$

Given $f_o = 2.0$ cm and $u_o = -2.5$ cm (object distance is negative by convention):

$$\frac{1}{2.0} = \frac{1}{v_o} - \frac{1}{-2.5}$$$$\frac{1}{2.0} = \frac{1}{v_o} + \frac{1}{2.5}$$$$\frac{1}{v_o} = \frac{1}{2.0} - \frac{1}{2.5} = \frac{2.5 - 2.0}{2.0 \times 2.5} = \frac{0.5}{5.0} = \frac{1}{10}$$ $$v_o = 10 \text{ cm}$$

The magnification produced by the objective lens ($m_o$) is:

$$m_o = \frac{v_o}{u_o} = \frac{10}{-2.5} = -4$$

Now, for the eyepiece, the final image is formed at the least distance of distinct vision ($D = 25$ cm). The image formed by the objective acts as the object for the eyepiece. Let the object distance for the eyepiece be $u_e$ and the image distance be $v_e = -D = -25$ cm. The focal length of the eyepiece is $f_e = 5.0$ cm. Using the lens formula for the eyepiece:

$$\frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e}$$$$\frac{1}{5.0} = \frac{1}{-25} - \frac{1}{u_e}$$$$\frac{1}{u_e} = -\frac{1}{25} - \frac{1}{5.0} = -\frac{1}{25} - \frac{5}{25} = -\frac{6}{25}$$ $$u_e = -\frac{25}{6} \text{ cm}$$

The magnification produced by the eyepiece ($m_e$) when the final image is at D is:

$$m_e = 1 + \frac{D}{f_e} = 1 + \frac{25}{5.0} = 1 + 5 = 6$$

The total magnifying power ($M$) of the compound microscope is the product of the magnifications of the objective and the eyepiece:

$$M = m_o \times m_e = (-4) \times 6 = -24$$

However, the options provided are in terms of magnitude. Let’s re-examine the eyepiece magnification formula when the image is at D. It is indeed $1 + \frac{D}{f_e}$.

Let’s double-check the calculations.

Objective lens: $u_o = -2.5$ cm, $f_o = 2.0$ cm $\frac{1}{v_o} = \frac{1}{f_o} + \frac{1}{u_o} = \frac{1}{2} - \frac{1}{2.5} = \frac{2.5 - 2}{5} = \frac{0.5}{5} = 0.1$ $v_o = 10$ cm $m_o = \frac{v_o}{|u_o|} = \frac{10}{2.5} = 4$ (magnitude)

Eyepiece: $f_e = 5$ cm, $D = 25$ cm Magnification of eyepiece when final image is at D: $m_e = 1 + \frac{D}{f_e} = 1 + \frac{25}{5} = 1 + 5 = 6$

Total magnifying power $M = m_o \times m_e = 4 \times 6 = 24$.

There seems to be a slight discrepancy with the provided options. Let’s review the problem statement and my calculations once more.

Ah, I see a potential point. The distance between the image formed by the objective and the eyepiece needs to be considered if it’s not explicitly stated that the image formed by the objective is at the focal point of the eyepiece for normal adjustment. However, the question mentions the final image is at the least distance of distinct vision, which implies a specific setup.

Let’s re-calculate $u_e$ for the eyepiece: $\frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e}$ $\frac{1}{5} = \frac{1}{-25} - \frac{1}{u_e}$ $\frac{1}{u_e} = -\frac{1}{25} - \frac{1}{5} = -\frac{1 + 5}{25} = -\frac{6}{25}$ $u_e = -\frac{25}{6}$ cm

Magnification of eyepiece $m_e = \frac{v_e}{u_e} = \frac{-25}{-25/6} = 6$.

Magnification of objective $m_o = \frac{v_o}{u_o} = \frac{10}{-2.5} = -4$.

Total magnification $M = m_o \times m_e = (-4) \times 6 = -24$. The magnitude is 24, which is not among the options.

Let’s consider the case where the image formed by the objective is very close to the eyepiece’s focal point, leading to a large angular magnification by the eyepiece, approximately $\frac{D}{f_e}$.

If we use the formula $M \approx -\frac{v_o}{u_o} \left(1 + \frac{D}{f_e}\right)$, we get $M \approx -\frac{10}{-2.5} \left(1 + \frac{25}{5}\right) = 4 \times 6 = 24$.

It seems there might be an issue with the provided options or a subtle aspect I’m overlooking. Let me think about alternative approaches or common approximations used in such problems.

One common approximation for the magnifying power of a compound microscope when the final image is at D is $M \approx -\frac{L}{f_o} \left(1 + \frac{D}{f_e}\right)$, where $L$ is the tube length (distance between the second focal point of the objective and the first focal point of the eyepiece). We don’t have $L$ here.

Let’s go back to the fundamental definitions without approximations.

$m_o = \frac{v_o}{u_o} = \frac{10}{-2.5} = -4$

For the eyepiece, the object distance is $u_e = -\frac{25}{6}$ cm, and $f_e = 5$ cm. The angular magnification of the eyepiece is $m_e = \frac{D}{|u_e|} = \frac{25}{25/6} = 6$.

The total magnifying power is $M = m_o \times m_e = -4 \times 6 = -24$. The magnitude is 24.

Given the options, there might be a slight variation in how magnifying power is defined (sometimes magnitude is considered). If we consider the magnitudes, the closest value isn’t directly present.

Let’s re-read the question carefully to ensure no detail was missed. The setup seems standard.

Could there be a calculation error? $\frac{1}{v_o} = \frac{1}{2} - \frac{1}{2.5} = 0.5 - 0.4 = 0.1 \implies v_o = 10$ cm. Correct. $m_o = -\frac{10}{2.5} = -4$. Correct. $\frac{1}{u_e} = \frac{1}{f_e} - \frac{1}{v_e} = \frac{1}{5} - \frac{1}{-25} = \frac{1}{5} + \frac{1}{25} = \frac{5+1}{25} = \frac{6}{25} \implies u_e = -\frac{25}{6}$ cm. Correct. $m_e = \frac{v_e}{u_e} = \frac{-25}{-25/6} = 6$. Correct. $M = m_o \times m_e = -4 \times 6 = -24$.

If we consider the magnitude, it’s 24, which is not an option. Let me consider a possible approximation where the image by the objective is assumed to be very far, and the eyepiece acts as a simple magnifier.

If the final image is at infinity (normal adjustment), $m_e = \frac{D}{f_e} = \frac{25}{5} = 5$. Then $M = m_o \times m_e = -4 \times 5 = -20$ (magnitude 20, not an option).

Let’s assume there might be a slight error in the options provided in a real exam scenario, or a specific convention being followed. If we had to choose the closest option based on our calculation of 24, option (2) 25 would be the nearest. However, this is speculative.

Let’s try to work backward from the options to see if any lead to a consistent scenario. This is generally not advisable but can help identify potential misunderstandings.

If $M = 25$, and $m_e = 1 + \frac{25}{5} = 6$, then $m_o = \frac{25}{6} = \frac{v_o}{2.5} \implies v_o = \frac{25 \times 2.5}{6} \approx 10.4$ cm. Using lens formula for objective: $\frac{1}{2} = \frac{1}{10.4} - \frac{1}{-2.5} = \frac{1}{10.4} + \frac{1}{2.5} \approx 0.096 + 0.4 = 0.496$, which is close to $0.5$. This suggests option (2) might be intended, possibly with slight rounding in values.

Given the analysis, and acknowledging a potential discrepancy or need for approximation relative to the options, the closest answer based on our detailed calculation is (2) 25.

Therefore, the closest answer is (2) 25.