Error Analysis in Concave Mirror Optical Bench Experiment

To find which of the given data sets cannot be obtained from the experiment, we use the mirror formula for a concave mirror:
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
where $u$ is the object distance, $v$ is the image distance, and $f$ is the focal length of the mirror.

Given that $f = 24\text{ cm}$, the theoretically expected image distance $v_{\text{actual}}$ for a given object distance $u$ is given by:
$v_{\text{actual}} = \frac{u f}{u - f} = \frac{24u}{u - 24}$

The maximum permissible error in the location of the image is $\Delta v_{\text{max}} = 0.2\text{ cm}$. Therefore, any correctly recorded value of $v$ must satisfy:
$|v - v_{\text{actual}}| \le 0.2\text{ cm}$

Additionally, the length of the optical bench is $1.5\text{ m} = 150\text{ cm}$, so the total distance between the object and the image on the bench must not exceed $150\text{ cm}$ (i.e., $u + v \le 150\text{ cm}$).

Now, we check each of the recorded sets of $(u, v)$:

1. For $(u, v) = (42\text{ cm}, 56\text{ cm})$:
   $v_{\text{actual}} = \frac{42 \times 24}{42 - 24} = \frac{1008}{18} = 56.0\text{ cm}$
   The difference is:
   $|v - v_{\text{actual}}| = |56 - 56.0| = 0\text{ cm} \le 0.2\text{ cm}$
   Also, $u + v = 42 + 56 = 98\text{ cm} \le 150\text{ cm}$.
   This data set is correctly recorded.

2. For $(u, v) = (48\text{ cm}, 48\text{ cm})$:
   $v_{\text{actual}} = \frac{48 \times 24}{48 - 24} = \frac{1152}{24} = 48.0\text{ cm}$
   The difference is:
   $|v - v_{\text{actual}}| = |48 - 48.0| = 0\text{ cm} \le 0.2\text{ cm}$
   Also, $u + v = 48 + 48 = 96\text{ cm} \le 150\text{ cm}$.
   This data set is correctly recorded.

3. For $(u, v) = (60\text{ cm}, 40\text{ cm})$:
   $v_{\text{actual}} = \frac{60 \times 24}{60 - 24} = \frac{1440}{36} = 40.0\text{ cm}$
   The difference is:
   $|v - v_{\text{actual}}| = |40 - 40.0| = 0\text{ cm} \le 0.2\text{ cm}$
   This data set is correctly recorded.

4. For $(u, v) = (66\text{ cm}, 33\text{ cm})$:
   $v_{\text{actual}} = \frac{66 \times 24}{66 - 24} = \frac{1584}{42} \approx 37.71\text{ cm}$
   The difference is:
   $|v - v_{\text{actual}}| = |33 - 37.71| = 4.71\text{ cm} > 0.2\text{ cm}$
   Since the error exceeds the maximum possible error of $0.2\text{ cm}$, this data set cannot come from the experiment.

5. For $(u, v) = (78\text{ cm}, 39\text{ cm})$:
   $v_{\text{actual}} = \frac{78 \times 24}{78 - 24} = \frac{1872}{54} \approx 34.67\text{ cm}$
   The difference is:
   $|v - v_{\text{actual}}| = |39 - 34.67| = 4.33\text{ cm} > 0.2\text{ cm}$
   Since the error exceeds the maximum possible error of $0.2\text{ cm}$, this data set cannot come from the experiment.

Therefore, the data sets that cannot come from the experiment and are incorrectly recorded are $(66, 33)$ and $(78, 39)$.

Correct Options: C and D