To calculate the refractive index \( n \) of the glass using the angle of minimum deviation \( D \) of a triangular prism, we can use the formula:
\[ n = \frac{\sin\left(\frac{A + D}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
where:
- \( A \) is the angle of the prism,
- \( D \) is the angle of minimum deviation.
In this case, you've provided the angle of minimum deviation as \( D = 42.6° \) (assuming the unit is degrees). But it appears you might have written \( 42.6\pi \), which looks like a misunderstanding because typically the angle in this context would be given in degrees or radians, not in a multiple of \( \pi \).
Assuming \( D = 42.6° \):
To use this formula, we also need the angle of the prism \( A \). Since the angle of the prism is not provided, we will proceed assuming a common value for \( A \) (which is typically 60° for many practical prisms). However, if you have a specific prism angle, please provide it.
Let’s calculate the refractive index for different values of \( A \):
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Assuming \( A = 60° \): \[ n = \frac{\sin\left(\frac{60° + 42.6°}{2}\right)}{\sin\left(\frac{60°}{2}\right)} \] \[ n = \frac{\sin(51.3°)}{\sin(30°)} \] \[ n = \frac{0.779}{0.5} = 1.558 \]
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If \( A \) is another value, let me know, and I can provide the calculation for that.
Thus, under the assumption of a \( 60° \) prism, the refractive index of the glass would be approximately \( 1.558 \).